' Handbook of

FINANCIAL

ALCULATIONS

'and

PHYSICAL

MEASUREMENTS

agriculture Handbook No. 230

U.S. DEPARTMENT OF AGRICULTURE

Economie Research Service

-arm Economies Division
•^j^aKMfc"

Modem American farming is a complex business operation. As

farms get bigger and investments higher, more and more figuring is

required for determining the costs and returns of the farm business and

for such things as depreciation. Social Security, credit, life insurance,

retirement, and estate planning.

Today's farmer needs to be able to figure the interest rates charged

by competing sellers and creditors, the size of installment required to

pay off a debt at a specified interest rate over a given term, survivorship

benefits payable to dependents under Social Security if he dies before

retirement, and his own retirement income if he lives. He needs also to

be able to convert life insurance proceeds into installment income for

dependents in order to know whether Social Security and present life

insurance will provide adequate incomes for them. And because financial

computations are based on physical measurements, he n e e d s to know

such things as how to estimate acreages, yields in the field, and bushels

in the bin.

This report is an attempt to make these and many other calculations

easier for farmers and for county agents, country bankers. State and

Federal employees, and others whom farmers call upon to advise and

assist them. The calculations presented are those most frequently asked

of Department of Agriculture and other public employees. The material

is presented in question form with detailed solutions, so that the reader

may solve similar problems with the help of the tables included in the

report.

Washington, D. C. April 1962

For sale by the Superintendent of Documents, U.S. Government Printing Office, Washington 25, D.C. - Price 35 cents
CONTENTS

Page

FINANCIAL CALCULATIONS 1

General Formula for Calculating Short-Term Interest Rates 1

Bank loans 2

Small loans 3

Installment purchas es 4

Payment of Long-Term Loans by Regular Installments

(Amortization) 4

Use of table 1 (principal and interest paid per $1 borrowed) 5

Depreciation 9

Straight-line method 9

Sum-of-years-digits method 9

Accumulation of Specified Sum by Regular Deposits 10

Use of tables 2 and 3(payments required to accumulate $1) 10

Life Insurance 14

Payments '^certain** with no life contingency 17

Payments involving a (single) life contingency 17

Joint life and two-thirds survivorship annuity 18

Social Security 21

Survivorship benefits 25

Retirement benefits 26

Lump -Sum Deposits 27

^'Life*^ of deposit from which regular monthly income

is withdrawn 27

Lump sum used to provide income for specified period 28

Planning Life Insurance 30

PHYSICAL MEASUREMENTS 36

Estimating Yields Per Acre in the Field 36

Corn 36

Cotton 36
CONTENTS --Continued

"page

_ _ 37

Acreage " ^

Rectangular fields

Triangular fields '^'

Opposite sides parallel (parallelogram) 38

Two sides parallel 38

No two sides parallel 38

More than four sides 39

Irregular boundaries 39

Volume (Cubic Contents) 41

Shelled corn or grain in bin 41

Ear corn in crib (shelled basis) 41

Piled grain 43

Corn silage in round silo 44

Corn silage in trench silo 44

Hay in mow 45

Hay in stack 45

Gallons of water in tank or cistern 46

Moisture content of corn 46

Conversion of bushels by volume to bushels by weight 47

Estimating Quantities of Building Materials Needed 48

Board feet 48

Concrete 48

Rafter length 48

Wallpaper . 48

Joists (16 inches on center) 49

Allowance for waste 49

Paint 50

APPENDIX ' 50

Try this way of solving some simple problems 50

Conversion factors 52

Measurement equivalents and formulas 53

II
FARMERS» HANDBOOK OF FINANCIAL CALCULATIONS

AND PHYSICAL MEASUREMENTS

Prepared by Ralph R. Botts, Agricultural Economist

Farm Economics Division, Economic Research Service

FINANCIAL CALCULATIONS

General Formula for Calculating Short-Term Interest Rates

An understanding of how to figure interest costs and rates could save you hundreds of

dollars over a period of years. The following formula and problem solutions will

help you determine whether lower interest rates are charged by one lender or seller

than by another.

2 X number of payment periods in 1 yeari' x interest—'

mterest = ^^r j-,

^^IQ balance—' x (number of payments—' +1)

Expressed more compactly, the formula is:

2ml

r = B (n+1)

_!/ Denoted by m in the compact formula. Regardless of the number of payments

you make, when payments are made monthly, use 12 here; if made semimonthly, use

24; if every 2 months, use 6; if weekly, use 52, etc.

2/ Denoted by I in the compact formula. Time price minus cash price, or amount

you pay back less the amount you receive in the case of a loan.

3/ Denoted by B in the compact formula. Cash price minus downpayment, or

amount you get in the case of a loan.

4/ Denoted by n in the compact formula.
Bank loans

Example #1: You borrow $100 from a bank for 6 months. The bank»s discount

rate is 4 percent a year, so it deducts $2 from your loan and gives you $98.

You are to repay the $100 in 6 months. What interest rate is being charged?

Solution: Since repayment is at the end of 6 months, there are two such payment periods in one year. Here m = 2. The interest charge I is $2. The

balance B is the $98 you get. You repay the loan in one lump sum, so n = 1

and n + 1 = 2.

Then r = "^-^^^-^ = 0.0408, or 4.1 percent.

Example #2: You borrow $1,000 from a bank that charges a discount rate of 5

percent a year. You are to repay the loan in 3 years. The bank deducts $150

from the loan (or 3 x 0.05 x 1,000) and gives you the remaining $850. You are

to pay back $1,000 in 3 years. What interest rate is being charged?

Solution: Since the $1,000 is to be repaid in 3 years, one year represents 1/3

of the payment period. Here m = 1/3. Only one payment is to be made, so

n = 1 and n + 1 = 2. I = $150 and B = $850.

rr.. 2x1/3x150 2x1x150 ^ ^.oo c o +

Th^^ ^ = 850 X 2 = 850x3x2 = ^'^^«^^ ^^ ^'^ P^^^^^^'

Example #3: You borrow $100 from a bank that charges 6 percent discount.

You are to repay the loan in 18 months. The bank deducts $9 from the loan

(or 1.5 X 0.06 X 100) and gives you $91. What interest rate is being charged?

Solution: Here m = 2/3, since a year represents only 12/18 or 2/3 of the

payment period. I = $9, B = $91, and n = l, sonH-l=2.

Then r = ^x2/3x9 ^ 2 x 2 x 9 0.0659, or 6.6 percent.

91 x2 91x3x2

Example #4: A bank advertises its "add-on" interest rate for new-car loans

at 5 percent. On a $1,000 loan paid back over 2 years, it would add $100

(or 2 X 0.05 X 1,000) to the amount of the loan. The borrower would pay the

$1,100 in 24 payments, so his monthly payment would be $45.83 (or 1,100 f 24).

What interest rate is being charged?

Solution: Here I = 1,000 x 2 x 0.05 = $100; B = $1,000; n = 24, n+ 1 = 25; and

m = 12. If payments are to be made monthly, m is always 12. If payments

are to be made weekly, m = 52; if semimonthly, m = 24; if quarterly, m = 4,

and so on.

T^^^ ^ = ^Ai^^"" o^r" = -096, or 9.6 percent. 1,000 X 25
B^sairñ'M»'#§•?-"-Y-oia-tee-rí-ow $1,200 from your bank to pay cash for a used car.

The bank's advertised interest rate is 4 percent. You are to repay the loan

in monthly installments spread over 18 months. How much are your monthly

payments? What is the annual interest rate?

Solution: Interest I = 1.5 x 0.04 x 1,200 = $72. So the loan would be for $1,272

(or 1,200 +72). B = 1,200; n+1 = 19. Your monthly payments are $70.67 (or

1,272 í 18). The annual interest rate r is 7.6 percent, computed as follows:

2 X 12 X 72

^= 1,200x19 '- '"''' °^ 7-6 percent.

The computed rate is almost twice the advertised rate, because the loan is

repaid by installments rather than by a lump-sum payment at the end of the

time.

Example #6: You borrow $1,200 from your bank and repay the loan at $70.67

per month for 18 months. What interest rate is being charged?

Solution: The interest charge I is $72.06, or the amount you pay back less the

amount you received, i.e., (18 x 70.67) - 1,200 is equal to 72.06. The balance

B is the amount received, or $1,200.

X 12 X 72.(

1.200 X 19

„, 2 X 12 X 72.06 r^ nn^o la Then r = — —— = 0.0758, or 7.6 percent.

Small loans

Example #1: You borrow $250 from a finance company. You are to repay the

loan in 11 semimonthly installments of $25 each (on the 1st and 15th of the

month). What interest rate is being charged?

Solution: Here the interest charge I = (11 x $25) - $250 or $25. Stated in words,

it is the amount you pay back less the amount you get. Here m = 24, since payments are to be made twice monthly and there are 24 such periods in one year.

Here n = 11 and n+1 = 12.

Then r = ^ ^^l'^ ^,1^ = 4/10 = 0.4 - 0.40 = 40 percent.

¿DO X 1¿

Example #2: You borrow $100. Your repayment schedule calls for 6 installments of $20 each, made every 2 months. What interest rate is being charged?

Solution: Here m = 6, since there are 6 bi-monthly periods in one year. The

time price or amount repaid is $120 (or 6 x $20). The interest charge I is the

difference between the amount paid back and the amount received, or $20 (which

is $120 - $100). The balance B is the $100 received from the lender.

Then r = ^ ^ ^ ^ ^° = 12/35 = 0.343, or 34.3 percent.

100 X 7
Installment purchases

Example #1: You can buy a refrigerator for $200 cash or for $20 down and 8

payments of-$25 each every 2 months. If you buy it on time, what interest

rate is being charged?

Solution: The interest charge I is $20 or the difference between the time

price and the cash price.

Time price = $20 + (8 x $25) = $220

Less cash price 200

Interest charge I 20

The balance B is the cash price minus the downpayment, or $200 - $20, which

is $180. m = 6; n = 8 and n+ 1 =9.

Then r = ^ ^ ^ ^ ^^ = .148, or 14.8 percent. 180x9

Example #2: You can buy a TV set for $200 cash or on time for $10 down and

5 monthly payments of $39 each. What interest rate is being charged if you

buy on tinae?

Solution: The interest charge I is the time price less the cash price, or $10 +

(5 x $39) - $200, which is $5. The balance B is the cash price minus the

downpayment, or $200 - $10, which is $190.

2x12x5 Then r = -— — = 0.105, or 10.5 percent.

Example #3: You are considering making a time purchase from a mail-order

house. Its catalog advertises that on a balance (after downpayment) of from

$180 to $200, the company will add on $25 as a service or time charge. You

get 18 months to pay the time price.

The article you are interested in is listed at $230, with $30 down. What

interest rate is being charged if you buy on time?

Solution: Your balance is $200 (or 230 - 30). You would have to pay back $225

(or $200 plus the time charge of $25), so each of your 18 monthly payments

would be $12.50 (which is 225 - 18). Here m = 12; n = 18 and n + 1 = 19. The

interest charge I is $25.

^^^"^ "* ^ ^'ool^'lf ^ ^'^^^' ""^ ^^'^ percent.

Payment of Long-Term Loans by Regular Installments (Amortization)

In buying a farm or additional land, a long-term 1 o an may be necessary The

foUowmg problems and their solutions explain how to determine the amount of a

periodic payment that will amortize (pay off) a debt in a given time ata given interest

rate, what your total payments will amount to, and how much of this total will be

interest or the cost of the credit extended to you.
Hge^of Table 1 (principal and interest paid per $1 borrowed)

Example #1: After a downpayment on farmland, your balance is $10,000. It is

to be amortized over 10 years by regular annual payments. The annual interest

rate is 6 percent.

a) How much will you repay lender?

b) What will be your annual payment?

c) How much interest will you pay?

Solution:

a) 10,000 X 1.358680 = $13,586.80. The 1.358680 is found in the 6-percent

column of table 1, opposite 10 annual payments.

b) 13,586.80^ 10 = $1,358.68 per year.

c) 13,586.80 - 10,000 = $3,586.80 interest.

Example #2: Same as #1 except your payments are to be made semiannually.

Solution:

a) 10,000 X 1.344314 = $13,443.14. The 1.344314 is found in the 6-percent

colunrm of table 1 opposite 20 semiannual payments.

b) 13,433.14 ^ 20 = $672.16 semiannually.

c) 13,443.14- 10,000 = $3,443.14 interest.

Example #3: Same as #1 except your payments are to be made quarterly.

Solution:

a) 10,000 X 1.337084 = $13,370.84. The 1.337084 is found in the 6-percent

column of table 1 opposite 40 quarterly periods.

b) 13,370.84 ^ 40 = $334.27 per quarter.

c) 13,370.84 - 10,000 = $3,370.84 interest.

Example #4: Same as #1 except your payments are to be made monthly.

Solution:

a) 10,000 X 1.332246 = $13,322.46. The 1.332246 is found in the 6-percent

column of table 1 opposite 120 monthly periods.

b) 13,322.46 4 120 = $111.02 per month.

c) 13,322.46 - 10,000 = $3,322.46 interest.
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Month

End of month

Interest

duel/

MonthlypaymentReduction of

principal

Balance

outstanding

Dollars Dollars

At time of loan1st

2d

3d

118th

119th

120th

50.00

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2/49.39

1.10

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111.02

111.02

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61.33

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109.92

110.47

Dollars

10,000.00

9,938.98

9,877.65

9,816.02

220.39

110.47

0

Total -/ 3,322.46 13,322.46 10,000.00

1^1 At 0.005 or one-half of 1 percent per month (since 0.06 t12 = 0.005).

2/ 9,877.65 x 0.005 = 49.39.

3/ 111.02 - 49.39 = 61.63.

4/ Due to rounding, columns may not add exactly to the totals

shown.

Annual payment $87,19

Reduction of

principal

Amortization of 20-year $1,000 6-percent loan

by equal annual payments of |87,19

While the annual payment remains the same, the part applicable to interest

declines, and the part applicable to reduction of the debt increases (see the

amortization schedule above).
Depreciation

A farmer who does not charge depreciation as a cost in his farming operation is

overstating his profits. In a way, he is "living off his principal." The depreciation

of a machine, for example, is just as much an operating cost as the gasoline used to

run it. And if he does not charge depreciation, he is paying more income taxes than

he should pay. The following problems and their solutions explain how depreciation

is calculated by two common methods.

Straight-line method

Annual

depreciation

cost - salvage

years of life

Example: A tractor costs $3,200. You estimate that it has a probable life of

12 years and a salvage value of $200 at the end of that time. What is the annual

depreciation on the machine? What is its carrying value at the end of 4 years?

At the end of 6 years ?

Solution:

Annual depreciation = —^ 200

12

3,000

12 = $250.

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At end of 4

At end of 6

years = 3,200 -

years = 3,200 -

(4 X

(6 X

250)

250)

= $2,200.

= $1,700.

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depreciation

Carrying

value!/

n

Dollars

250

250

250

250

250

250

250

250

250

Dollars

3,200

2,950

2,700

2,450

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_1 / Value at which you would carry the tractor as an asset on

your financial statement at end of year shown.

Sum-of-years-digits method

Example: Same as above.
Solution: Sum of years digits = 1 + 2 + 3 + . . . + 11 + 12 = 78.

Also 12 X ^ t ^^ = 6 X 13 = 78.

Depreciation

1st year = (12/78) x (3,200 - 200) = 461.54.

2d year = (11/78) x 3,000 = (3,000/78) x 11 = 38.462 x 11 = 423.08.

3d year = 38.462 x 10 = 384.62.

4th year = 38.462 x 9 = 346.16.

6th year - 38.462 x 7 = 269.23.

llthyear = 38.462x2 =(3,000/78)x 2 = (2/78) x 3,000 = 76.92.

12thyear = 38.462 x 1 = (1/78) x 3,000 = 38.46.

Year Annual . Carrying

depreciation . value 1/

n

; Dollars Dollars

: 3,200.00

: 461.54 2,738.46

: 423.08 2,315.38

384.62 1,930.76

346.16 1,584.60

307.70 1,276.90

269.23 1,007.67

230.77 776.90

192.31 584.59

153.85 430.74

115.39 315.35

76.92 238.43

38.46 2/ 199.97

11— -

9 _ _

Q

Art

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7

8 :

qy .

10 :

11 :

12 .

Total ; 3/ 3,000.03

_!/ Value at which you would carry the tractor as an asset on

your financial statement at end of year shown.

_2/ Does not equal $200 because of rounding,

_3/ Does not equal $3,000 because of rounding.

Accumulation of Specified Sum by Regular Deposits

If you save up and buy for cash, instead of buying on time, your income will go a lot

further. You'll not only save finance charges, but your savings will also earn

mterest. So, if you can. save first and buy later. The following problems and their

solutions show how to determine the periodic payment required to build up a "sinking

fund of a certam sum in a given time period at a given interest rate.

Use of tables 2 and 3 (payments required to accumulate $1)

If the number of months between deposits is the same duration as the interestconversion period (table 2).

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Example #1: You will need $400 for a new roof on youi- ßäi'li lu lU'fëàrS.

How much must you put aside annually, beginning now, in order to have ^

the $400 in 10 years, if your savings earn 4 percent compounded annuallySolution: Your deposits are made annually. Interest also is compounded

annually. Both are 12-month periods; therefore, use table 2.

400 X 0.08008745 = 32.03. You must deposit $32.03 per year,

beginning now. After 10 years, you will have deposited $320.30

(or 10 X 32.03). Interest of $79.70 will bring the total up to $400.

The 0.08008745 is found in the 4-percent column of table 2

opposite 10 periods (years in this case).

Example #2: Same as #1 except your savings accumulate at 4 percent

compounded semiannually and you make semiannual deposits. How much

would you need to deposit every 6 months, beginning now, in order to

have the $400 at the end of 10 years?

Solution: The payment period and the interest-conversion period are both

6 months. Therefore, use table 2.

400 X 0.04034972 = 16.14. You would need to deposit $16.14

every 6 months, beginning now. At the end of 10 years, you

will have deposited $322.80 (or 20 x 16.14), but interest of

$77.20 brings your savings up to $400.

The 0.040M^72 is found in the 2-percent column of table 2

opposite 20 payment periods. There are 20 payment periods

and interest is at 2 percent per payment (semiannual) period.

Example #3: You want to save $1,800 in 6 years to send a son to college.

How much would you need to put aside quarterly, beginning now, if your

money earns 4 percent compounded quarterly?

Solution: 1,800 x 0.03670641 = 66.07. In 6 years (24 periods) you deposit

$1,585.68 (or 24 x 66.07). Interest of $214.32 brings the total up to the

$1,800 required. The 0.03670641 is found in the 1 percent (or 0.04 + 4)

column of table 2 opposite 24 quarterly periods.

Payments made more frequently or less frequently than interest is compounded

(table 2 and 3).

Example #4: You want to accumulate $1,000 in 5 years to send your son

to college. You can get 4 percent compounded quarterly on savings. How

much would you need to deposit monthly, beginning now?

Solution: 1,000 x 0.04496566 x 0.33443953 = $15.04 per month. The

0.04496566 is found in the 1-percent column of table 2, opposite 20 periods.

The quarterly rate is 1 percent, and there are 20 quarterly periods in the

5 years. But there is only one month between deposits, whereas there are

3 months in the interest-conversion period. Therefore, a correction factor of 0.33443953 must be used. It appears in the "1/3" column of table 3,

opposite 1 percent.

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Example #5: Same as above except you want to make semiannual, instead

of monthly, deposits. How much would you need to deposit every 6 monx

beginning now?

Solution- 1 000 X 0.04496566 x 1.99009901 = $89.49 semiannually. The

1 99009901 \s found in the column headed "2", opposite 1 percent,^ m

table 3 There are 6 months between deposits and only 3 months m the

interes't-conversion period. Therefore, toe ratio 6/3 = 2; we look opposite

1 percent in the column designated as "2" m table 3.

Example #6: Same as #4 except you want to make your deposits every 3

months. How much would these deposits be, beginning now?

Solution: 1,000 x 0.04496566 = $44.97 quarterly. No correction factor

from table 3 is needed, since the payment period and the interestconversion period are the same length (3 months). The 0.04496566 is

from the 1-percent column of table 2, opposite 20 quarterly periods.

This problem is handled from table 2 in the same way as examples #1,

#2, and #3.

Example #7: You want to save $1,800 in 6 years to send a son to college.

How much would you need to deposit annually, beginning now, if your

savings earn 4 percent compounded quarterly?

Solution: 1,800 x 0.03670641 x 3.94098521 = $260.39 per year. The

0.03670641 is found in the 1-percent column of table 2, opposite 24

interest periods. The interest rate is 1 percent quarterly (0.04 f 4),

and there are 24 quarters in the 6-year period.

Then, since there are 12 months between deposits and only 3 months

between interest conversions (12/3 = 4), look in the "4" column of

table 3 for a correction factor. Use 3.94098521, which is found in the

4" column of table 3 opposite the 1-percent interest rate. !» ylH

Life Insurance

Tables 4 and 5 are useful in planning life insurance to fit your needs. From these

tables, you can determine how far a given amount of life insurance proceeds would

go in providing a monthly income for your widow and other dependents in case of

your death.

An option under which you may elect a monthly income for your widow, instead of a

lump-sum payment of the face amount, or under which she may so elect if you do

not choose one of the optional modes of settlement, is included in every policy.

Therefore, each policy includes a table showing the money incomes applicable to

various income periods from which a choice may be made. The monthly incomes

shown in this table are based on the guaranteed interest rate stated in the policy.

Companies may pay more than the guaranteed rate when warranted by current investment earnings. To be on the conservative side, the monthly incomes shown in

the above-mentioned table (the same as one of the columns of table 4) or computed

from table 5, which are based on a guaranteed interest rate, should be used for

planning purposes.

14
Table^üj^^-Lilfe mÉémiñée^ierm certain): Monthly income to beneficiary per $1,000

of policy face amount by (1) interest rate, and (2) length of income period

selected 1/

Income period

(years)

Interest rate specified in policy

2 percent 2 1/2 percent 3 percent

Monthly income per $1,000 of face amount

3--

4_.

5--

6--

!-

8--

9--

10.

II'

12

13

14'

15

16

17

18

19

20

21

22

23

24

25

Doll ars

28 59

21 65

17 49

14 72

12 74

11 25

10 10

9 18

8.42

7 .80

7 .26

6 .81

6 .42

6 .07

5 .77

5 .50

5 .26

5 .04

4 .85

4 .67

4 .51

4 .36

4 .22

Dollars

28.79

21.86

17.70

14.93

12,95

11.47

10.32

9.39

8.64

8.02

7.49

7.03

6.64

6.30

6.00

5.73

5.49

5.27

5.08

4.90

4.74

4.60

4.46

Dollars

28.99

22.06

17.91

15.14

13.16

11.68

10.53

9.61

8.86

8.24

7.71

7.26

6.87

6.53

6.23

5.96

5.73

5.51

5.32

5.15

4.99

4.84

4.71

1/ Payments continue to estate if beneficiary dies before end of selected period.

Calculations are for an "annuity due," that is, one that is payable at the beginning of

each month following death of insured person.

15
Table 5.--Life insurance monthly income to beneficiary per $1,000 of policy face amount, by

type of settlement option and by age of widow at time of death of insured husband_l/

35-

36-

37-

38-

39-

40-

41-

42-

43-

44-

45-

46-

47-

48-

49-

50-

51-

52-

53-

54-

55-

56-

57-

58-

59-

60-

61-

62-

63-

64-

65-

66-

67-

68-

69-

70-

Age of widow

(years)

Monthly income per $1,000 of insurance

Life annuity with

10 years certain^/

Dollars

3.27

3.31

3.36

3.40

3.45

3.50

3.56

3.61

67

73

79

86

93

00

08

16

4.24

4.32

4.42

4.51

4.61

4.72

4.83

4.95

5.07

5.20

5.34

5.48

5.64

5.80

5.97

6.15

6.34

6.54

6.75

6.97

Dollars

3.26

3.30

3.34

3.39

3.43

3.48

3.53

3.59

3.64

3.70

3.76

3.82

3.88

3.95

4.02

4-09

4.17

4.25

4.33

4.42

4.50

4.60

4.69

4.79

4.90

5.01

5.12

5.23

5.35

5.48

5.61

5.74

5.87

6.01

6.16

6.30

11 Calculations are based on 1937 Standard Annuitants' Mortality Table and a 2 1/2 percent

interest rate, with female life "aged back" 6 years, and with no "loading" for operating

expenses. These are "annuities due, " that is, payments are at beginning of each monthly

period. For male beneficiaries, subtract 5 years from ages shown. For example, the monthly

incomes for a female aged 45 are applicable to a male aged 40.

_2/ Payments cease upon death of beneficiary.

^/ If beneficiary dies before the end of 10 years, payments continue to the estate for the

remainder of the original 10-year period. If beneficiary dies after 10 years, payments cease

at that time.

16
PayníHhfcfí ^^ rev^^^-^^^ w^^S^o life contingency'

Example #1: Suppose you have $5,000 worth of life insurance. You want to know

what monthly income it would provide for, say, 10 years in case of your death.

Solution: Look in the section of your policy dealing with settlement options, and

find the interest rate applicable to proceeds retained by the company. Many

policies issued since 1948 are at 2 1/2 percent. Suppose yours is also 2 1/2

percent.

Then, looking in the 2 1/2 percent column of table 4, opposite 10 years, we see

that $1,000 would provide $9.39 per month, so $5,000 would provide $46.95 per

month (or 5 x 9.39). The table in your policy shows the same figures as the

"2 1/2" percent column of table 4.

Example #2: You decide that $46.95 is too little. You want to know how far the

$5,000 would go at $100 a month.

Solution: Divide 100 by 5 to get $20 per month. Each $1,000 of insurance must

provide $20 a month. Looking for $20 in the 2 1/2 percent column of table 4,

we find that it falls between $21.86 for 4 years and $17.70 for 5 years; so the

$5,000 of insurance proceeds would provide $100 a month for between 4 and 5

years. _5/

Payments involving a (single) life contingency

Straight life annuity

Example #1: You have $10,000 of life insurance and want to use it to

provide the maximum monthly income for your widow for the remainder

of her life. She is now aged 55. If you should die now, how much would

she get per month under a life annuity elected by you or her?

Solution: Each $1,000 of proceeds would provide $4.61 per month, so your

$10,000 would provide $46.10 per month (or 10 x 4.61). The 4.61 is found

in the "Life annuity" column of table 5, opposite widow's age 55.

Under this option, if she died soon, payments would stop. But if she lived

to very old age, her payments would continue to the end of her life.

Life annuity with 10 years of payments guaranteed

Example #2: Under the same conditions, how much would be provided by

a life annuity with 10 years of payments certain?

5/ For greater accuracy, interpolation is required, as follows:

21.86 - 20 :, i_i8Ë = 0.447. Therefore, a more precise answer would be 4.447

21.86 - 17.70 . 4.16

years or 4 years and 5 months.

17
Solution: Under this option, $4.50 per month per $i,OÜÜ would be payable

to a widow aged 55. The $10,000 of face amount therefore would provide

$45 (or 10 X 4.50). The 4.50 is found in the "Life annuity with 10 years

certain" column of table 5, opposite widow^s age 55.

In comparison with the straight life annuity, used in example #1, this

option would reduce the monthly income by $1.10 per month. Under it,

the company would make payments to the estate for the remainder of the

10-year guaranteed period (beginning at date of your death) if your widow

should die before the end of the 10 years. If she should live beyond the

10 years, her payments would continue to the end of her life, as in the

case of the straight life annuity in example #1 above.

Example #3: How much life insurance would be required to provide your

widow with a monthly income of $150 under a "lO-years-certain" life

annuity, as explained in #2 above? She is now aged 65.

Solution: At age 65, each $1,000 would buy $5.61 per month. The 5.61 is

found in the "Life annuity with 10 years certain" column of table 5,

opposite age 65. Then 150/5.61 = 26.738, so it would take $26,738 of life

insurance to provide the monthly income of $150. You could get a policy

for, say, $27,000. If you already have a policy for $10,000, you would

need about $17,000 more insurance to provide the desired monthly income

for your widow.

Under a straight life annuity (no payments certain), the $26,738 of proceeds would buy $159.63 per month (or 26.738 x 5.97). (See table 5.) The

term certain therefore reduces the monthly payments by $9.63 (difference

between $159.63 and $150). A term certain reduces the monthly payments

much more at the older than at the younger ages (fig. 1).

Joint life and two-thirds survivorship annuity

Based on joint life expectancies, table 6 can be used to convert a lump sum, for

example, life insurance cash values, into monthly (annuity) incomes for a

husband and wife during their remaining years. The survivor gets only twothirds as much as the couple received while both were alive. Two examples of

use of the table are given.

Under this type of annuity, the amounts shown in table 6 are payable while both

husband and wife are alive. After one (either one) dies, the survivor gets twothirds of the amounts shown in table 6. The last survivor receives her (or his)

annuity until death. There is no refund for early death of both annuitants.

Example #1: Suppose you are aged 65; your wife is 62. You are both

eligible for OASDI (social security) benefits. You have an endowment-atage-65 policy which matures soon for $5,000. _6/ You want to know how

_6/ The $5,000 might also represent the cash value of a longer term endowment, or

the cash value of a limited pay or an ordinary life policy, for a greater face amount

than $5,000.

18
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19
much monthly income the $5,000 would provide under a joint lite and^

two-thirds survivorship annuity, as described above. You want to

supplement your social security money with this additional income.

Solution: (5,000/1,000) x 5.61 = $28.05 per month. After one of you

dies, the survivor gets $18.70 per month. The 5.61 is found in row

"65" for age of husband, and under column headed "Husband older

than wife by 3 years."

Example #2: A son would be willing to take over the operation of your

farm. However, it represents your life savings, so you must have a

retirement income from it. If it is agreed that the value of the farm is

$60,000 (after deducting for use of the dwelling by the parents during

their remaining years), how much income per month would this sale

value provide under a joint life and two-thirds survivorship annuity?

The father is 62, his wife is 58.

Solution: (60,000/1,000) x 5.11 = $306.60 per month.

In accordance with the agreement, the son would pay his parents $306.60

per month as long as both were alive. After the death of one, the son

would pay the survivor $204.40 per month. The son would not be relieved

of his debt until both parents were dead. If one or both lived long, he

would still be required to meet his payments; if both died early, his debt

would be paid off. All children should understand how the agreement works.

In effect, the son would be assuming the responsibilities of a life insurance

company. This procedure does provide, however, a way in which a farm

can be passed along to a son (or son-in-law) as a going operation.

MONTHLY INCOME PROVIDED WIDOW

DOLLARS

*

Life annuity with 10 years of

payments guaranteed^

45 50 55 60 65 70

AGE OF WIDOW (YEARS)^

*PER $1,000 OF INSURANCE O PAYMENTS CONTINUE FOR 10 YEARS, EVEN THOUGH WIDOW DIES IN MEANTIkiE

A AT TIME OF HUSBAND'S DEATH

U. S.DEPARTMENT OF AGRICULTURE NEC. ERS 211-61(10) ECONOMIC RESEARCH SERVICE

Figure 1

20
Social Security

Unless you can estimate or find out from your local Social Security office (1) how

much monthly income your widow and dependent children would get from Social

Security if you died, you cannot (2) estimate whether your present life insurance and

other income is adequate for them. And you can't plan your own retirement unless

(3) you know about what your Social Security retirement income will be if you live.

Table 7 is used in solving problems under (1), and tables 7 and 8 are used in solving

problems under (3). Problems coming under (2) are solved in other parts of the

report.

The benefits payable under social security may change

from time to time as new legislation is passed. The

payments shown or calculated in this report were in

effect after the 1961 amendments.

In all life insurance and estate planning, survivorship and retirement benefits payable under social security are considered first. A person is "currently" insured if

he worked in covered employment at least half of the time during the 3 years before

he dies. He is "fully" insured if he worked at least one-fourth of the quarters since

1950. If he is fully insured at the time of his death, his family is entitled to survivorship benefits if there are children then under age 18. Moreover, if he is fully

insured, he and his wife (and any children then under 18) are entitled to retirement

benefits at age 62.

Both survivorship and retirement benefits depend on average earnings under social

security, as evidenced by income-tax returns. For those qualifying, these monthly

payments cannot be less than certain established minimums. There is also a family

maximum.

Average

monthly

earnings

gross earnings 1951 to date of death or retirement

(after dropout of up to 5 years of lowest income)

elapsed months since January 1, 1951, or subsequent

"entry" (after dropout of up to 60 months)

There are 5 steps involved in this calculation:

1. You need your gross earnings figures for every year since 1950.

2. Drop your lowest 5 years.

3. Add up your remaining income to get total income for these years.

4. Add up elapsed months in the years for which you use the earnings.

5. Divide the figure obtained in step 3 by the figure obtained in step 4.

If you will take your earnings record by individual

years to your local social security office, someone

there can furnish you with survivor-benefit figures

or even a retirement income estimate based on it.

21
Table 7.--Social Security: Reduced PIA's, survivorship benefits, and retirement benefits ^^^

husband 64, wife 63, by average monthly wages

Retirement

: Reduced

Survivorship benefits benefits

Average : PIA Widow 62 or . Widow under 62 with this

monthly

wages

1/ over, no

children

. number of children under

age 18

Family

maximum^/

Husband 64,

wife 631/

1.6875 X (1) under age 18

1 2

: ^^^ :

(2) (3) (4) (5) ; (6)

: Dollars Dollars Dollars Dollars Dollars Dollars

67 or less : 32.00 40.00 60.00 60.00 60.00 54.00

68-69 : 32.80 40.00 61.60 61.50 61.50 55.40

7 0- : 33.60

: 34.40

40.00

40.00

63.00

64.60

63.00

64.50

63.00

64.50

56.70

71-72 58.10

73-74 35.20 40.00 66.00 66.00 66.00 59.40

75-76 36.00 40.00 67.60 67.50 67.50 60.80

77-78 36.80 40.00 69.00 69.00 69.00 62.10

79-80 37.60 40.00 70.60 70.50 70.50 63.50

Q1 _ __ oi~ — 38.40 40.00 72.00 72.00 72.00 64.80

82-83 39.20 40.50 73.60 73.50 73.50 66.20

84-85 : 40.00 41.30 75.00 75.00 75.00 67.50

86-87 : 40.80 42.10 76.60 76.50 76.50 68.90

88-89 : 41.60 42.90 78.00 78.00 78.00 70.20

qn-- -- • 42.40 43.80 79.60 79.50 79.50 71.60

91-92 : 43.20 44.60 81.00 81.00 81.00 72.90

93-94 . 44.00 45.40 82.60 82.50 82.50 74.30

95-96 :

Q7 .

44.80

45.60

46.20

47.10

84.00

85.60

84.00

85.50

84-00

85.50

75.60

77.00

98-99 : 46.40 47.90 87.00 87.00 87.00 78.30

100-101 :

1 09. „_ .

47.20

48.00

48.70

49.50

88.60

90.00

88.50

90.00

88.50 79.70

90-00 81.00

103-104 : 48.80 50.40 91.60 91.50 91.50 82.40

105-106 : 49.60 51.20 93.00 93.00 93.00 83.70

1 07 50.40 52.00 94.60 94.50 94.50 85.10

108-109 : 51.20 52.80 96.00 96.00 96-00 86.40

110-113 : 52.00 53.70 97.60 97.50 97.50 87.80

114-118 : 52.80 54.50 99.00 99.00 99.00 89.10 119-122 : 53.60 55.30 100.60 100.50 100.50 90.50 123-127 : 54.40 56.10 102.00 102.00 102.00 91.80 128-132 : 55.20 57-00 103.60 105.60 105.60 93.20 133-136 : 56.00 57.80 105-00 108.90 108.80 94.50 137-141 : 56.80 58.60 106.60 112.80 112.80 95.90 142-146 : 57.60 59.40 108.00 117.00 116.80 97.20 147-150 : 58.40 60.30 109.60 120.00 120.00 98.60 151-155 : 59.20 61.10 111.00 124.20 124.00 99.90 156-160 : 60.00 61.90 112.60 128.10 128.00 101.30 161-164 : 60.80 62.70 114.00 131.40 131.20 102.60 165-169 : 61.60 63.60 115.60 135.30 135.20 104.00

See footnotes at end of table.

22
Tablè^'^BSSiiii'gitniidl Oe^,lfífl!y!^^educed PIA's, survivorship benefits, and retirement benefits for

husband 64, wife 63, by average monthly wages --Continued

Reduced

PIA

II

Survivorship benefits Retirement

benefits

Average

monthly

wages :

Widow 62 or

over, no

children

under age 18

Widow under 62 with this

number of children under

age 18

Family

maximum.2/

Husband 64,

wife 631/

1 2

1.6875 X (1)

(1) (2) (3) (4) (5) (6)

Dollars Dollars Dollars Dollars Dollars Dollars

170-174 62.40 64.40 117.00 139.20 139.20 105.30

175-178 63.20 65.20 118.60 142.50 142.40 106.70

179-183 64.00 66.00 120.00 146.40 146.40 108.00

184-188 64.80 66.90 121.60 150.60 150.40 . 109.40

189-193 65.60 67.70 123.00 154.50 154.40 110.70

194-197 66.40 68.50 124.60 157.80 157.60 112.10

198-202 67.20 69.30 126.00 161.70 161.60 113.40

203-207 68.00 70.20 127.60 165.60 165.60 114.80

208-211 68.80 71.00 129.00 168.90 168.80 116.10

212-216 69.60 71.80 130.60 172.80 172.80 117.50

217-221 70.40 72.60 132.00 177.00 176.80 118.80

222-225 71.20 73.50 133.60 180.00 180.00 120.20

226-230 72.00 74.30 135.00 184.20 184.00 121.50

231-235 72.80 75.10 136.60 188.10 188.00 122.90

236-239 73.60 75.90 138.00 191.40 191.20 124.20

240-244 74.40 76.80 139.60 195.30 195.20 125.60

245-249 75.20 77.60 141.00 199.20 199.20 126.90

250-253 76.00 78.40 142.60 202.50 202.40 128.30

254-258 76.80 79.20 144.00 206.40 206.40 129.60

259-263 77.60 80.10 145.60 210.60 210.40 131.00

264-267 78.40 80.90 147.00 213.60 213.60 132.30

268-272 79.20 81.70 148.60 217.80 217.60 133.70

273-277 80.00 82.50 150.00 221.70 221.60 135.00

278-281 80.80 83.40 151.60 225.00 224.80 136.40

282-286 81.60 84.20 153.00 228.90 228.80 137.70

287-291 82.40 85.00 154.60 231.90 232.80 139.10

292-295 83.20 85.80 156.00 234.00 236.00 140.40

296-300 84.00 86.70 157.60 236.40 240.00 141.80

301-305 84.80 87.50 159.00 238.50 244.00 143.10

306-309 85.60 88.30 160.60 240.90 247.20 144.50

310-314 86.40 89.10 162.00 243.00 251.20 145.80

315-319 87.20 90.00 163.60 245.40 254.00 147.20

320-323 88.00 90.80 165.00 247.50 254.00 148.50

324-328 : 88.80 91.60 166.60 249.90 254.00 149.90

329-333 89.60 92.40 168.00 252.00 254.00 151.20

334-337 : 90.40 93.30 169.60 254.10 254.00 152.60

338-342 : 91.20 94.10 171.00 254.10 254.00 153.90

343-347 : 92.00 94.90 172.60 254.10 254.00 155.30

See footnotes at end of table.

23
Table 7.—Social Security: Reduced PIA's, survivorship benefits, and retirement benefits for

husband 64, wife 63, by average monthly wages --Continued

: Reduced

: PIA

' Il

Survivorship benefits

Retirement

benefits

Average

monthly

wages

Widow 62 or

over, no

children

under age 18

'. Widow under 62 with this

number of children under

age 18

Family

maximum^/

Husband 64,

wife 631/

1.6875 X (1)

1 2

(1) . (2) (3) (4) (5) (6)

Dollars Dollars Dollars Dollars Dollars Dollars

348-351 92.80 95.70 174.00 254.10 254.00 156.60

352-356 93.60 96.60 175.60 254.10 254.00 158.00

357-361 : 94.40 97.40 177.00 254.10 254.00 159.30

362-365 95.20 98.20 178.60 254.10 254.00 160.70

366-370 : 96.00 99.00 180.00 254.10 254.00 162.00

371-375 : 96.80 99.90 181.60 254.10 254.00 163.40

376-379 : 97.60 100.70 183.00 254.10 254.00 164.70

380-384 : 98.40 101.50 184.60 254.10 254.00 166.10

385-389 : 99.20 102.30 186.00 254.10 254.00 167.40

390-393 : 100.00 103.20 187.60 254.10 254.00 168.80

394-398 : 100.80 104.00 189.00 254.10 254.00 170.10

399-400 : 101.60 104.80 190.60 254.10 254.00 171.50

_!/ A farmer who retires at age 62, and whose wife is less than 62, would receive amounts

shown in this column until she became 62.

2/ Amounts in this column applicable to a widow less than 62, with 3 or more children under

age 18.

_3/ Because of rounding of figures applicable to individuals, a few figures in this column are

10 or 20 cents less than the actual amounts would be. The age-combination factor "1.6875"

comes from table 8.

Table 8.--Social security retirement income of couple retiring between ages 62 and 65, per

dollar of husband's reduced PIA for age 62

Age of

husband

62.

63.

64.

65.

62

Dollars

1.468750

1.552083

1.635417

1.718750

Age of wife

63

Dollars

1.520833

1.604167

1.687500

1.770833

64

Dollars

1.572917

1.656250

1.739583

1.822917

65

Dollars

1.625000

1.708333

K791667

1.875000

24
The bäS'lä föi^"all survivor benefits to your family in case of your death, and to you

and your wife if you both live to retirement, is the primary insurance amount (PIA),

which is based on your average monthly earnings under social security. 7/ The

maximum income that could be counted was changed in a recent year from $4,200 to

$4,800. The year of change is not important, as you can get a statement of your

social security earnings from the Social Security Administration that will show all

earnings that can be counted, in accordance with the maximums that were in effect

when the earnings were made.

Table 7 shows reduced PIA's, survivorship benefits, and the retirement benefits

applicable to a husband aged 64 and wife aged 63, by average monthly wages. More

attention will be given to retirement benefits for other husband-and-wife age combinations later.

Survivorship benefits

Example #1:- A widow is aged 35 at the time of her husband's death; their

only child is aged 13. The husband was fully insured. His covered earnings

averaged $284 per month. What monthly benefit will his family receive?

How long will they receive these benefits?

Solution: For average monthly earnings of $284, the family benefit for a

widow with one child under 18 is $153 (from table 7). The family will

receive this $153 per month for 5 years, or until the child is 18. Then all

payments will stop. At 62, if she has not remarried, the widow will begin

to receive payments of $84.20 per month. 8/

Example #2: A widow is aged 40 at the time of her husband's death; their

children are aged 6 and 10. At his death, the husband was fully insured.

His covered earnings averaged $258 per month. What monthly benefit will

his family receive? For how long?

Solution: For average monthly wages of $258, the benefit for a widow with

2 dependent children is $206.40 per month (found in column 4, table 7,

opposite average monthly wages of $258).

When the older child is 18, the younger child will be 14. Then payments to

the widow and younger child will drop to $144 per month (found in column 3,

table 7, opposite average monthly wages of $258).

When this younger child is 18, the widow will be 52. Then all social security

payments will stop until she reaches 62. At that time, she will start receiving checks for $79.20 (found in column 2, table 7, opposite average

monthly wages of $258). She will continue receiving them until she dies or

remarries.

7^/ In this report, the husband's primary insurance amount applicable to age 62

will be referred to as the "reduced PIA."

8/ The $84.20 is found in column (3) of table 7, opposite average monthly wages

of $284 per month.

25
Retirement benefits

Example #1: A farmer retires at 62, at which time his wife is aged 60. His

covered earnings averaged $261 per month. What will be their combined

social security income?

Solution: Nothing is payable for the wife until she reaches 62. The payment

for the husband is his reduced PIA applicable to average monthly wages of

$261. This figure (from column 1 of table 7) is $77.60 per month. When she

becomes 62, their check will be increased to $114 (or 1.46875 x 77.60). 9/

The payment for him does not change. He continues to receive 100 percent

of his reduced PIA, since his checks started at age 62. But the payment for

her amounts to 0.46875 x 77.60 or $36.40. _10/ The total of $114 (or 77.60 +

36.40) is 1.46875 x his reduced PIA (after rounding). The 1.46875 is given in

table 8 for husband aged 62 (at entry, even though he is now 64) and for wife

aged 62 at entry.

Example #2: A farmer retires at age 64; his wife is 63. If his covered earnings since 1950 averaged $340 per month after a dropout of up to 5 years of

lowest income, what will be their combined social security retirement income

per month?

Solution: His reduced PIA applicable to age 62 would be $91.20 (from column

1 of table 7 opposite average monthly wages of $340). Then 1.6875 x 91.20 =

$153.90. The 1.6875 is found in table 8 opposite age 64 for age of husband,

under column headed "63" for age of wife.

The family social security retirement income for a husband retiring at age

64, with wife then aged 63 and no children under 18, is shown, by average

monthly wages, in column 6, table 7. Therefore, the $153.90 also might have

been found in column 6, table 7, opposite average monthly wages of $340.

Example #3: Same as #2 except let's assume that this farmer is 5 months past

his 64th birthday at the time of his retirement, and that his wife is 9 months

past her 63rd birthday at that time. What will be the couple's social security

retirement income?

Solution: Use the following formula for fractional ages between 62 and 65

inclusive —

Retired ^^ , T . Reduced couple's =

annuity PTA J. ±r^

(1 + N/144) + H (1 + n/108)

where N = number of months husband is past his 62d birthday,

and n = number of months his wife is past her 62d birthday.

9/ 1.46875 X 77.60 = 113.975. It is then rounded upward to nearest 10 cents that

is, to $114.

JLO/ 0.46875 X 77.60 = 36.375. It is then rounded upward to nearest 10 cents that

is, to $36.40.

26
T'ú't' aÜT"S?ample';"'tíie figures are as follows:

Retired

couplers = 91.20 jd + 29/144) + "^ d + 21/108]]

annuity -, r

= 91.20 [(1 + 0.2014)+ 32 (1 + 0.1944]]

= 91.20 fl.2014 + 0.46875(1.1944")]

= 91.20 ]^1.2014 + 0.55991

= 91.20 X 1.7613

= 160.63. Use $160.70.

Of course this result is a little higher than the $153.90 obtained in example #2.

There it is assumed that the husband is exactly 64 at retirement, and that his

wife is exactly 63 at that time.

Lump-Sum Deposits

Here, problems are solved that involve depositing a lump sum at interest and withdrawing a regular monthly income from the account. Balances continue to draw

interest. The problem is to find out how long the fund will last. For example, a

farmer may want to find out how long his life insurance proceeds would last his

widow if she deposited them in a savings and loan association paying 4 1/4 percent

compounded semiannually and withdrew $100 a month from the account. Savings and

loan associations usually pay higher interest (dividend) rates than those used in the

annuity calculations of life insurance companies.

"Life" of deposit from which regular monthly income is withdrawn

Example #1: If $5,000 of proceeds from a life insurance policy were deposited

in a savings and loan association paying 4 percent compounded quarterly, and

if 2 percent of the $5,000 (or $100) were withdrawn monthly from the account,

how long would the fund last?

Solution: 4.56 years, or 4 years and 6.72 months. The 4.56 years is found in

the "2-percent" column of table 9 opposite "4 percent compounded quarterly."

After the 54th withdrawal (4 years and 6 months), there would be about $72 left

in the fund (or 0.72 x 100) for withdrawal the following month.

Example #2: If only $50 per month is withdrawn from the fund described in

example #1 above, how long will the fund last?

Solution: The fund will last 10.14 years. The $50 is 1 percent of the $5,000

deposit, so the answer is found in the "l-percent" column of table 9 opposite

"4 percent compounded quarterly."

The fractional part of a year amounts to 1.68 months, so the full $50 per month

could be withdrawn for 121 months, at which time there would be about $34 left

(0.68 X 50 = 34).

27
Table 9.--"Life" of a lump-sum deposit which earns interest while a regular monthlyincome is withdrawn from it

[ If this percentage of the lump-sum deposit

[ is withdrawn each month

Interest earned on deposit,

and frequencj^ of compounding ; 0.65 ; 0.75 ; i ; 2 ; 3 ; 4

; Fund will last this long:

4 percent, compounded —

Years Years Years Years Years Years

17.86 14.62 ,10.11 4.56 2.95 2.18

17.94 14.67 10.13 4.56 2.95 2.18

17.99 14.70 10.14 4.56 2.95 2.18

18.01 14.72 10.15 4.57 2.95 2.18

20.20 16.06 10.73 4.67 2.99 2.20

20.38 16.16 10.77 4.68 2.99 2.20

20.47 16.22 10.79 4.68 3.00 2.20

20.53 16.25 10.80 4.68 3.00 2.20

23.71 17.97 11.44 4.79 3.04 2.23

24.13 18.17 11.52 4.80 3.04 2.23

24.29 18.28 11.56 4.80 3.05 2.23

24.50 18.36 11.58 4.81 3.05 2.23

iinnuaiiy ~ —

oemiannuaiiy

^í^uarieriy—

ivioniniy ~ ~

5 percent, compounded — :

-Tirmuaiiy " .

oClllJLdl 11 lUcLiJ.y " — — — — — — — — — ^

C^l 1 Í3 K*'h O T»T T7" _. — — __ — — — _ ______

IX/Tr^-n+VilTr- _ •

6 percent, compounded — :

xXlUl U.cLL±y *" """ ______ _

QcifYTi Q'nT*moTTTr___ ___ ____ _

v^ucLi Ltii ly — :

Mon+Vil-vr 1VHJ11L11±Y ~.

*

Lump sum used to provide income for specified period

Example #1: Suppose your wife is aged 50 and your youngest child is aged 13.

When this child is 18, your wife will be 55. If you should die in the meantime,

your widow would not be eligible for a social security survivorship annuity

between the ages of 55 and 62. During this 7-year period, she would have to

rely on income from life insurance proceeds or from other sources.

If you had $10,000 of life insurance, and your widow invested that amount of

proceeds at 4 percent compounded semiannually, how much could she withdraw

from her account each month and still be sure that, with interest, the fund

would last for 7 years ?

Solution: 10,000 x 0.01365 = $136.50 per month. The 1.365 percent is found in

the column headed "4 percent compounded semiannually" of table 10, opposite

"7" years. If she deposited the $10,000 at age 55, her fund would last until she

was 62.

Example #2: How much life insurance would you need, under the situation

described in example #1, to provide your widow with $200 per month during

this 7-year period?

28
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29
Solution: 200 -r 0.01365 = 14,652. You would need $14,652 of life insurance, so

you would have to take out another policy for about $5,000. The 1.365 percent

is found in table 10 under the column headed "4 percent compounded semiannually, " opposite "7" years.

Example #3: If your widow financed, at 5 percent payable monthly over a 5-yea

period, a balance of $5,000 on a well-secured first mortgage, what monthly

amount would she get as principal and interest?

Solution: 0.01886 x 5,000 = $94.30 per month. The 1.886 percent is found in the

column headed "5 percent compounded quarterly" of table 10, opposite 5 years.

Using the percentages in the "compounded quarterly" columns of table 10 for

monthly withdrawals creates very little error. JJ^/ The withdrawal or compounding of interest by collecting or paying it monthly, rather than quarterly,

adds very little to the amount of the periodic withdrawal or payment.

Planning Life Insurance

Here is an example of life insurance programing (planning). A farmer who has

dependents needs occasionally to make at least a rough estimate of the monthly income his dependents would receive if he should die. The tables and calculations may

be helpful in making such an estimate.

The usual procedure in determining whether life insurance is adequate is to assume

that the provider dies now. What income would dependents receive after the expenses

of the last illness were met and after debts were paid?

Example: A farmer aged 37 has a wife 36 and two children aged 13 and 10. He

has $20,000 of life insurance. He owes a balance of $5,000 on a mortgage. He

has no "mortgage redemption" life insurance. How might he "program" his

life insurance to see whether it is adequate?

Solution: He sends a card to the Social Security Administration requesting a

statement of his "earnings and quarters of coverage." On the card, he enters

his account number, date of birth, signature, and (printed) name and address.

In response to his inquiry, he receives a Statement showing that he has earned

33 quarters of coverage, and that he is fully insured and has qualified for minimum retirement benefits. The statement also shows that his covered earnings '

since 1950 amounted to $24,173. This total includes nonfarm wages earned

before he came under social security as a farm operator in 1955. By years or

periods, the statement shows the following information:

JL^/ For example, using table 1, the exact monthly payment (or withdrawal) is

calculated as follows:

5,000 X 1.132274 5,661.37 ^^, ,, ^, 5-7-Ï2 " ""^O " $94.36. The 1.132274 is found in

the "5 percent" column of table 1, opposite 60 monthly payments.

30
Period Earnings

1951-56 $12,728

1957 4,200

1958 3,123

1959 4.122

Total for 9 years--- $24,173

The statement does not show a breakdown of his earnings by years for the

6-year period 1951-56, so unless he has kept copies of his income-tax returns

or a separate record of his reported social security earnings for these years,

he will not know the figures by individual years.

As only the earnings for the most recent 3 years are shown by years, the

grouped period will become longer and longer, so it becomes increasingly

necessary for a farmer to keep an earnings record of his own for programing

purposes in connection with the dropout of the lowest 5 years of earnings. Of

course, when he dies or retires, the calculations will be made for him from

more detailed records available to the Social Security Administration.

At this point, we know that this farmer is fully insured, because more than half

of the elapsed quarters since 1950 are covered. He has 33 quarters of a possible 36 (or 9 years x 4 quarters per year). Therefore, his family is entitled to

survivorship benefits if he dies and he and his wife are entitled to at least minimum retirement benefits. Moreover, if his earnings continue at about the same

level as in the past, their retirement benefits will be substantially more than

the minimum. But in a programing study, we assume death now of the breadwinner, so we are primarily interested in survivorship benefits to dependents.

In calculating average annual or monthly earnings for the period 1951-59, or for

any period for which the calculation is made, a dropout of the lowest 5 years is

allowed, as stated previously. For use in estimating his family's survivorship

inconae from time to time, this farmer should have copies of his income-tax

returns or a separate record of his reported earnings under social security.

For example, if he knew that in 1951 and 1953 he had very little or no nonfarm

earnings, he could drop these years out in computing his average annual earnings. This would raise his average above the figure that will now be calculated

if he had kept no records, does not remember any of the figures by years, and

must assume the average for each of the 6 years.

His average for 1951-56 was $2,121 (or 12,728/6), so he drops out 5 of these 6

years.

Average 24,173 - (5 x 2,121) 13,568

"^"^'^^y ~- 12 X (9 - 5) = -48-

wages

= 282.67 or $283.

For average monthly wages of $283, the widow and 2 children will get $228.90

per month (column 4, table 7, opposite average monthly wages of $283).

31
After the older child is 18, payments to the widow and younger child drop to

$153 per month (column 3, table 7, opposite average monthly wages of $283).

When the younger child is 18, the widow will be 44. As there no longer will be

any dependent children under 18, and she will be under age 62, payments to her

will stop. At age 62, she will start receiving checks for $84.20 (from column 2,

table 7, opposite average monthly wages of $283).

Period

1961-65-

1965-68-

1968-86-

1987 onYears

Age of family

Widow Younger

child

Older

child

Payments to - Amount

4 36-40 10-14

3 41-43 15-17

L8 44-61

62 on

13 - 17 Widow and both $228.90

children

Widow and 153.00

younger child

Widow 0

Widow 84.20

If $1,000 of the proceeds from life insurance were used for expenses of the last

illness, and if $5,000 were used to pay off the mortgage, that would leave

$14,000 to provide for the education of the children and for a monthly income for

the family. The widow will have an income froni social security until the

yoimger child is 18, at which time she will be 44, so let's see how much all of

the $14,000 would provide as a monthly income to fill the gap between 1968,

when the younger child will be 18, and 1986, when the widow (at 62) will begin

receiving her social security checks for $84.20.

Using a savings and loan association

By putting all of the $14,000 into one or more savings and loan

associations that pay 4 percent compounded quarterly, the fund

could be built up to $18,498 between 1961 and 1968._12/ The

widow would have that amount in her account when her social

security checks for $153 were stopped in 1968 because there

were no longer any children under 18.

How much could she withdraw monthly from her account, beginning in 1968, in order for the fund to last 18 years (from 1968

to 1986), at which time she would be aged 62 and could start

drawing her social security checks for $84.20 monthly?

Referring to table 10, we see that if she should draw out 0.65

percent (or 0.0065 as a decimal part) of the $18,498 monthly,

the fund would last the full 18 years. 13/

JL2/ 1.3213 X 14,000 = 18,498. The 1.3213 comes from the column of table 11 headed

"4 percent compounded quarterly, " opposite "7" years.

13/ The 0.65 percent is found in the column of table 10 headed "4 percent compounded quarterly, " opposite "18" years.

32
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33
This would be $120 per month (or 0.0065 x 18,498). After that, she

would start receiving her social security checks for $84.20, as illustrated below:

riod Years Age of

widow

Monthly income

Pe] Social

: Savings :

• and loan ; Total

Security- ; association;

Number Years Dollars Dollars Dollars

1961-

1965-

1968-

1986

-65--

-68--

-86--

on--

4

3

18

36-

41-

44-

62 (

40

43

61

228.90

153.00

0

84.20

0

0

120.00

0

228.90

153.00

120.00

84.20

Using a life insurance annuity

If the $18,498 in the widow's savings and loan association account (s)

in 1968 were used at that time to buy a life insurance annuity (no payments certain) at her attained age of 44, it would provide $64.50 per

month. ^/ However, this $64.50 would come in for the remainder of

her life, regardless of how long she lived.

If she lived beyond age 62, her total income would be $84.20 from

social security plus $64.50 from life insurance. She would get this

$148.70 per month as long as she lived (fig. 2). The social security

survivorship benefits and the life insurance proceeds would then be

as shown in the tabulation on page 35.

With this knowledge, this farmer probably would decide on more term

life insurance for a while. His existing program provides too little

for the widow during the "gap" period, and nothing for education of

the children. Later, after the couple is older, the term insurance

could be dropped. At the wife's older age, the remaining life insurance would buy for her a larger life annuity.

14/ Using the factor 3.73 from the "life annuity" column of table 5, opposite age

44, 3.73 X 18.498 = $69. But as we are assuming that the life annuity would not be

obtained as an option on her husband's life insurance policy, it would actually amount

to only $64.50 per month (or 0.935 x 69). Annuities not purchased through a settlement option on a life insurance policy are usually "loaded" by 6 1/2 percent for

company expenses.

34
Period

1961-65-

1965-68-

1968-86-

1986 onYears Age of

widow

Monthly income

Social

securityLife

insurance

Total

Number

4

3

18

Years

36-40

41-43

44-61

62 on

Dollars

228.90

153.00

0

84.20

Dollars

0

0

1/ 64.50

1/ 64.50

Dollars

228.90

153.00

1/ 64.50

1/ 148.70

_l/ Upon request, the insurance company could arrange the widow's life annuity so

she would get a constant monthly income from 1968 on. Under this arrangement, she

would get $96.46 from life insurance during the period 1968-86, and nothing from

social security. Then, from 1986 on, she would get $12.26 from life insurance and

the $84.20 from social security, making a total of $96.46 per month for the later as

well as the earlier period. These life-annuity calculations are based on the 1937

Standard Annuitant's Mortality Table and a 2 1/2 percent interest rate, with the

widow's age 6 years less than actual. They include a loading of 6 1/2 percent for

company expenses.

PROGRAMING OF WIDOW'S INCOME'

DOLLARS

180

120

60

L^

AGE OF WIDOW (YEARS)

J^FROM SOCIAL SECURITY AND LIFE INSURANCE, IF HUSBAND DIES WHEN WIFE IS 36

U. S. DEPARTMENT OF AGRICULTURE NEC. ERS 212-61(10) ECONOMIC RESEARCH SERVICE

Figure 2

35
PHYSICAL MEASUREMENTS

Estimating Yields Per Acre in the Field

Corn

Yield per acre in ^^^g counted on row of prescribed length (see below) x 100

70-pound baskets = ^^^-^^^ ^f ^^^^ in 70-pound (2-bushel) basket (see below)

of ear corn

If distance between rows is - Count the ears of corn on a row of this

length:

3 feet 0 inches

3 " 2 "

3 " 4 "

3 " 6 "

3 " 8 "

145 feet

137 "

131 "

124 "

119 "

Ear length (average)

10 inches and over

9 inches

8

7

6.

5

4

Number required to weight 70 pounds

80

90

100

130

180

225

260

Example: What is the yield in 70-pound (2-bushel)baskets of ear corn per acre

for a field in which the rows average 3 feet 4 inches apart?

Solution: Measure off 131 feet on a row and count the ears. Suppose there are

65, They average 7 inches in length.

„, 65 X 100 en X, 1 X

i hen = 50 baskets per acre. 130 ^

From the 50 (2-bushel) baskets of ear corn would be obtained

50 bushels of shelled corn.

Cotton

Yield per

acre (lint)

bolls (or empty burrs) counted on prescribed distance

(see below) x 0.5 15/

15/ There are about 200 bolls of seed cotton to 1 pound of lint cotton. The footage

measurements in the tabulation are for 1/100 of an acre. Multiplying the boll count

by 100 puts the count on an acreage basis; then dividing the result by 200 gives pounds

of lint per acre. But 100/200 equals 1/2. so 0.5 is used in the computation

36
If distance between rows is -

3 feet 0 inches

3 " 3 ''

Q !l c !I

Count the bolls (or empty burrs) on

a row of this length:

145 feet

135 feet

125 feet

Example: What is the lint yield per acre on a field with rows planted 3 feet

3 inches apart?

Solution: Measure off 135 feet on a typical row and count the bolls. Suppose

there are 1,200.

Then 1,200 x 0,5 = 600 pounds of lint.

OR count the bolls (or empty burrs) on half the distances shown in

above table, and use the count as a poundage figure. In example:

Count of 600 bolls = 600 pounds of lint.

(Count made on row 67 1/2 feet long).

Acreage

1 acre = 43,560 square feet = 4,840 square yards.

Acres area in square feet area in square yea43,560 4.840

Rectangular fields

A _. length X width (in feet) _ length x width (in yards)

Acres - 43,560 " 4,840

Length

Width

Example: A field measures 3,242 by 1,624 feet. How many acres are there?

3,242 X 1,624 Solution:

43,560

120.9 acres.

Triangular fields

base X altitude Acres =

2 X 43,560

Base Base

37
Example: A triangular field has a base of 5,000 feet and an altitude of 3,000

feet. How many acres are in the field?

c, -, ^. 5,000 X 3,000 .„n o

Solution: —^ = 172.2 acres. 2 X 43,560

Opposite sides parallel (parallelogram)

Altitude

I

É

Base

base X altitude (in feet) _ base x altitude (in yards)

^''''^^ ^ 437560 ^ 4784Ô

Example: How many acres are in a field,

opposite sides parallel, for which the

base measures 4,840 feet and the altitude

measures 3,260 feet?

Solution:

4,840 X 3,260

43,560 = 370 acres.

B

\

Width

I

or

1613 1/3 X 1086 2/3 _ 4,840 x 3,260

4,840 3 X 3 X 4,840 370 acres.

Two sides parallel

Arreci - (A+B) x width

^^""^^ 2 X 43,560

Example: How many acres in a field with side A measuring 900 feet and

B measuring 1,200 feet, with a width of 600 feet?

9nln.Hor.. (900+ 1,200) X 600 bolution: ' ^^^ = 14.5 acres. 2 X 43,560

No two sides parallel

Divide into two triangles with common

base; find areas separately and add.

Example: Area of field in illustration.

Solution: Triangle A has an altitude of

150 feet; triangle B, 400 feet. The

common base measures 500 feet.

38
Acres =

500 X 150 _^ 500 X 400

2 X 43,560 2 x 43,560

500 X (150 + 400)

2 X 43,560

500 X 550

2 X 43,560 3.2 acres.

More than four sides

Divide into triangles, find area

of each; then add.

Example: Area of field in

illustration.

Solution:

^ç^^^g ^ (150 X 390) + (750 x 300) + (750 x 250)

2 X 43,560

^ 58,500 + 750(300+250) ^ 58,500 + (750 x 550)

87,120 87,120

^ 58,500 + 412,500 ^ 471,000

87,120 87,120 ^

=5.4 acres.

Irregular boundaries

Somewhere below the irregular boundary, lay off a straight line (AB) between

opposite ends of the field. Measure the length of this line._l^/ At regular

intervals along this line, measure perpendicular lines between AB and the

irregular side. Be sure these perpendiculars are uniformly spaced._17/ In the

example, these perpendiculars vary in length from 27 to 155 feet. The approximate area of that part of the field between AB and the irregular boundary is

then, (1/2 of the sum of the two end perpendiculars, plus the sum of all other

perpendiculars) multiplied by the uniform spacing between perpendiculars.

J^/ In the example, it is 250 feet.

17/ In example, eveiy 25 feet.

39
Example: Area of field in illustration.

All distances in feet

Solution:

Area between

AB and the

irregular side

Area of

remainder

of field

25 X

E100 +

. 2

27

+ 150 + 155 + 150 + 60 + 33 +

47 + 100 + 70 + 70|

= 25 X 898.5

= 22,462 square feet.

125 X 250 = 31,250 square feet.

The area of the remainder of the field is then added to the area of the

portion bounded by the irregular side, to determine the total area

The total area in the field is then divided by 43,560 to determine the

acreage:

22,462+ 31,250

43:560 = ^-2 ^^^^s40
Volume (Cubic Contents)

Shelled corn or grain in bin

Rectangular bin

Bushels = 0.8 X volume = 0.8 x length x width x average depth

Example: A crib 18 feet long and 12 feet wide is filled with shelled corn

to an average depth of 8 feet. How many bushels are there?

Solution: 0.8 x 18 x 12 x 8 = 1,382 bushels.

Round bin

Bushels = 0.6283 x diameter x diameter x average depthi^/

Example: A round crib 12 feet 5 inches in diameter is filled with shelled

corn to an average depth of 6 feet 4 inches. How many bushels are there?

Solution:

12 ft. 5 in. = 12.4 ft.

6 ft. 4 in. = 6.3 ft.

Then 0.6283 x 12.4 x 12.4 x 6.3 = 609 bushels.

Ear corn in crib (shelled basis)

If dry, bushels = 4/9 x volume.

If new, bushels = 4/10 x volume.

If damp, bushels = 4/11 x volume.

Rectangular crib

Volume = length x width x average depth.

Example: A crib 18 feet long and 12 feet wide is filled with new ear corn

to an average depth of 8 feet. How many bushels are there?

Solution: 4/10 x 18 x 12 x 8 = 691 bushels.

18/ The volume of a cylindrical bin is 22/7 x radius x radius x average depth.

Expressing the radius as diameter/2, one might express the volume as 22/28 x

diameter x diameter x average depth. The 22/28 is 0.7854. Multiplying by 0.8,

the fractional part of a bushel in 1 cubic foot, gives 0.6283.

41
Round crib

Volume = 0.7854 x diameter x diameter x average depth

Example: A crib 10 feet in diameter is filled with dr^ ear corn to a

depth of 6 feet 6 inches. How many bushels are there?

Solution: 4/9 x 0.7854 x 10 x 10 x 6.5 = 227 bushels.

(6 feet 6 inches = 6.5 feet.)

Alternative (weight) method for either rectangular or round cribs

Bushels of shelled pounds of corn (to nearest tenth)

corn (if ear corn = shelled from 5-pound sample of

were shelled) ear corn

cubic feet of earn corn in crib

^ 10

Example: How many bushels of shelled corn are there in a crib

containing 1,000 cubic feet of ear corn? (The ears are not filled

out to standard.)

Solution: 5 pounds of ear corn from the crib are shelled. The

shelled corn weighs 3.2 pounds.

Then 3.2 x ^'^^^ = 320 bushels.

If 10 pounds of ear corn are shelled, divide cubic feet by 20,

instead of 10. In this example:

6Ax^^^ = 320 bushels. 20

Deduction for studding (for greater accuracy)

Bushels to be deducted = number of studs x depth x figure

from table below.

Multiply by

Studs

For grain ; ^ ^^^ tr/r""- ^

(on snelled basis)

2 by 4 inches

2 by 6 inches

2 by 8 inches

0.044 0.022

.067 ,033

.089 .044

42
¡Shelled corn or grain

Example: A crib 18 by 12 feet contains shelled corn to a depth

of 8 feet. On page 41, it was found that the crib contains 1,382

bushels, without deducting for studding.

If this crib has thirty 2- by 4-inch studs, how many bushels

must be deducted from 1,382 to get net bushels which allow for

displacement of grain by the studding?

Solution: 30 x 8 x 0.044 = 11 bushels.

Then 1,382 - 11 = 1,371 bushels net.

Ear corn (shelled basis)

Example: A crib 10 feet in diameter is filled with dry ear corn

to a depth of 6 feet 6 inches. Without allowing for studding, we

found on page 42 that the crib contained 227 bushels of shelled

corn.

If the crib has twelve 2- by 6-inch studs, how many bushels

must be deducted from 227 to get net bushels?

Solution: 12 x 6.5 x 0.033 = 2.574, or about 3 bushels.

Then 227-3 = 224 bushels net.

Piled grain

Shelled corn, small grains, and so on

Bushels = 0.20944 x height x diameter of base x diameter of base.

Example: How many bushels of

wheat are there in a cone-shaped

stack 10 feet high and 30 feet

across the base?

Solution: 0.20944 x 10 x 30 x

30 = 1,885 bushels.

Ear corn (shelled basis)

Bushels = 0.10472 x height x

diameter of base x

diameter of base.

Example: How many bushels of shelled corn are there in a stack of

ear com 10 feet high and 30 feet across the base?

Solution: 0.10472 x 10 x 30 x 30 = 1/2 of 1,885 (see above example) =

942.5 bushels.

43
By volume, 1 cubic foot of ear corn makes about 0.4 bushel of shelled corn.

A 2-bushel basket of ear corn makes 1 bushel of shelled corn.

By weight, the 2-bushel basket weighs about 70 pounds. It contains 56 pounds

of shelled corn (1 bushel) and 14 pounds of cobs (also 1 bushel). Therefore,

by weight, the corn is about 80 percent grain and 20 percent cobs.

Corn silage in round silo

_ 0.7854 X depth of silage x diameter x diameter

Cubic feet per ton (see tabulation below)

Depth of

settled

silage

(feet)

: :: Depth of

: Cubic feet :: settled

: per ton :: silage

:: (feet)

: Cubic feet

: per ton

14

16 :

18 :

20 :

22 :

24 .

26 :

28 :

55.3 :: 30 :

54.5 :: 32 :

54.0 ::34 :

53.3 ::36 :

53.0 ::38 :

52.5 ::40 :

52.1 ::42 :

R1 7 . • /I /I - _

51.3

51.0

50.7

50.5

50.3

50.1

49.9

01.< ..14 : 49.7

/' ^ / / ///

ft

0)

Dieineter

Example: A silo 16 feet inside diameter has 30 feet of silage in it. How manytons are there?

o 1 .. 0.7854 x30xl6xl6 ^^n n .

Solution: —— = 117.6 tons. 51.3

Corn silage in trench silo

Tons = average width x length x depth of silage x tons per cubic foot

of depth (see tabulation below).

1

2

3

4

5

6

7

8

Depth of

settled

silage

(feet)

Tons per

cubic foot

Depth of

settled

silage

(feet)

0.00925

.00985

.01040

.01090

.01140

.01190

.01235

,01280

9--

loll12-

13-

14-

15-

Tons per

cubic foot

0.01320

.01365

.01405

.01445

.01490

.01530

.01565

44
Example: A trench silo measures 8 feet wide at the bottom and 12 feet wide

at the top. The silage averages 8 feet deep and the silo is 50 feet long.

How many tons are there ?

Solution: MX 50 X 8 X 0.0128 = 51.2 tons.

Hay in mow

Tons = volume f a figure ranging from 435 to 640 (see tabulation below,

which applies to both hay in mow and hay in stack).

Volume = length x width x average depth.

Example: Timothy hay, 2 months old, fills a mow 30 by 40 feet to an average

depth of 9 feet 9 inches. How many tons are in the mow?

Solution: ^Q^^Q^^'^^ ^ ^3.3 tons.

640

Kind of hay

Alfalfa--

Clover--

Timothy

Wild

Cubic feet per ton

If in stack

30-90 days

If in stack more

than 90 days

485

450 to 485

640

600

470

435 to 470

625

450

Hay in stack

1 Q /

T = 0-3 X length x width x overthrow-^^^^

A figure ranging from 435 to 640

(from tabulation above)

Example: Approximately how many tons of alfalfa hay are in a stack 26

by 23 feet with an overthrow of 48 feet? The hay was harvested more

than 3 months ago.

Solution: 0.3 X 26 X 23 X 48

470

= 18.3 tons.

19/ Overthrow is the distance in feet over the stack from the ground on one side to

the~ground on the other side.

45
Gallons of water in tank or cistern

Rectangular

Gallons = length x width x depth x 7,48 20/

Example: How many gallons in a tank 12 feet 2 inches by 6 feet

4 inches by 5 feet 3 inches?

Solution: 12 1/6x6 1/3x5 1/4x7.48

^hich is 73x19x21x7.48 __ 3^026 gallons.

6x3x4

Cylindrical

Gallons

22 X diameter x diameter x height x 1.87

Example: How many gallons in a tank 5 feet 4 inches in diameter

and 7 feet 8 inches high?

Solution: 22 X 5 1/3 X 5 1/3 x 7 2/3 x 1.87

^ 22x16x16x23x1.87 ^ -^^282 gallons,

7x3x3x3

Moisture content of corn

Following are

to net bushels

correction factors for converting gross bushels of ear corn

Moisture

content

(percent)

Moisture

Factor .. content

(percent

Factor .

\ Moisture

\ content

(percent)

Factor

15 or less

ifi

1 090 "99 - - 0.925 :

.910 :

.895 :

.880 :

.865 :

.850 :

.835 :

. 9Q_ 0.820

.805

.790

.775

.760

.745

.730

1 OÍR .. 9*^--

: ¿la

i D

17

i.UiO .. ^o — —

1 000 ••94

: 6\j

. Q 1 1 1

1 Q

Ji ,yJ\J\J ., Ci'± — — —

QfiR -9^ -

: oi

± o . Q9

19

Of)

Q70 .. 9R--

: 6 ó

. QQ

QRR ••97 --

: oo —

. Q A

91 Q40 -9«

: o4

: 00

20/ 1,728/231 = 7.48. Multiplying cubic feet by 1.728 converts to cubic inches

Then dividing by 231 gives gallons. There are 231 cubic inches in one gallon.

46
Example: In an earlier example (page 41 ), it was estimated that there were

691 bushels of shelled corn in a rectangular crib 18 feet long and 12 feet

wide, filled with new ear corn to an average depth of 8 feet. Suppose a

sample of it shows the corn to have a moisture content of 23.3 percent. How

many net bushels of shelled corn are there?

Solution: The 23.3 percent moisture content is rounded to 23 percent before

entering the above tabulation. Then 0.91 x 691 = 628.8, or 629 bushels.

The 91 percent comes from the tabulation given above.

Conversion of bushels by volume to bushels by weight

Test weight of

1 bushel

(pounds)

Wheat, soybeans,,

and beans

Flax, shelled corn,

grain sorghum,

and rye

Barley

Multiply bushels by volume by this figure

to obtain bushels by weight

64-

62-

60-

58-

56-

54-

52-

50-

48-

46-

44-

42-

40-

38-

36-

1.07

1.03

1.00

.97

.93

.90

.87

.83

.80

.77

.73

.70

.67

.63

.60

1.07

1.04

1.00

.96

.93

.89

.86

.82

.79

.75

.71

.68

.64

1.08

1.04

1.00

.96

.92

.87

.83

.79

.75

Example: After deducting for displacement of grain by studding, we found in

an earlier example (page 43) that there were 1,371 bushels of shelled corn in

a rectangular bin 18 feet long and 12 feet wide, filled to an average (leveled)

depth of 8 feet. If a bushel of the corn weighs 46 pounds by test, how many

standard-weight bushels are in the crib?

Solution: 0.82 x 1,371 = 1,124 bushels by weight. The 0.82 comes from the

above tabulation. Since there are 56 pounds of shelled corn in a standard

bushel, 46 f 56 = 0.82.

47
Estimating Quantities of Building Materials Needed

Board feet

Board feet = (number of pieces x nominal thickness in inches x nominal

width in inches x length in feet) 4 12.

Example: How many board feet are in nine 2 by 4's fourteen feet long?

Solution: — = 84 board feet.

Concrete

Volume = l^^g"^^ ^^ fe,e\ x width in feet x thickness in inches

27 X 12

= answer in cubic yards.

Example: How much concrete mix will it take to make a sidewalk 52 feet

long, 2 feet 3 inches wide, and 4 inches thick?

Q^in.+.-^v. 52 X 2 1/4 X 4 52 X 9 X 4 .^o i..

Solution: ^^ ^ ^^ = 27x4x12 = ' ^^^ ^^^^^ ^^^^^^

The 2 1/4 in numerator = 9/4.

Omit the 27 in denominator and answer is in cubic feet;

answer, 39 cubic feet.

Rafter length

Rafter length = square root of (1/4 of width squared + rise squared.)

Example: A building 32 feet wide (distance AB)

is to have a rise of 12 feet (distance OC). What ^

rafter length (distance AC) is required?

Solution: Rafter length = square root of (1/4 x

32 X 32) + (12 X 12). The square root of 256 +

144 = square root of 400 = 20 feet.

(Proof: 20 X 20 = 400.)

Wallpaper

Single rolls ... 18 inches wide and 24 feet long ... 36 square feet.

Double rolls ... 18 inches wide and 48 feet long ... 72 square feet.

1.1

Rolls required =

square feet in walls or ceiling - square feetl

^in openings (windows and doors) | 21

36 or 72 ... depending on whether you use single

or double rolls.

/

]A¡ Allows 10 percent for waste and for matching needs.

48
iüxample: A room 7 feet high is 13 feet wide and 18 feet long. It contains

3 doors, 6 feet 2 inches by 2 feet 8 inches; and 2 windows, 4 feet by 2 feet

6 inches. How many single rolls of wallpaper are required to cover the

walls? Double rolls to cover the ceiling?

Solution:

For the walls

(less openings) ... 1.1 [(2x7x13) + (2x7x18) -(3x61/6x2 2/3)

- (2x4x2.5)] 4- 36

= l.l|^182 + 252 - 49 1/3 - 20)] ^ 36

= (1.1 X 364 2/3) 4. 36 = 401.133/36

= 11.1, or 12 single rolls.

For the ceilings . . . 1.1(13x18) ^ 1.1 x 234 ^ 257.4

72 72 72

= 3.6, or 4 double rolls.

Joists (16 inches on center)

Number required = (0.75 x length) + 1.

Allow 4 to 6 inches at each end for bearing on wall.

Allowance for waste

Add to area

for waste

Sheathing:

Common:

Laid horizontally (without openings in wall) l/lO

For dwellings . . figure as though without openings 0

Laid diagonally 1/6

Tight:

Laid horizontally:

6 -inch boards 1/5

8-inch boards l/"^

10-inch boards 1/9

Laid diagonally:

6-inch boards 1/4

8-inch boards 1/6

10 -inch bo ards 1/8

Flooring:

3-inch matched 1/2

4-inch matched 1/3

6-inch matched 1/5

Siding:

Drop

Lap:

4 inches to weather 1/2

4 1/2 inches to weather 1/3

49

1/5
Example: How many board feet of 1 by 6-inch sheathing laid tight diagonally

are required to cover a wall 10 feet 3 inches by 18 feet 4 inches?

Solution: 1.25 x 10 1/4 x 18 1/3 = 1.25 xll xA^

1.25 X 41 X 55 2818.75

4x3 12

= about 235 board feet.

Multiplying by 1.25 adds 25 percent for waste.

Paint

Area to be painted

Exterior:

Siding

Interior:

Woodwork

Plaster walls or ceiling

Gallons per 100 square feet

First coat Second coat

0.22

.25

.20

0.20

.17

.18

Example: How many gallons of paint woiñd be required to put a first coat of

paint on the plastered walls of the room used in example on page 49? It is

18 feet long, 13 feet wide, and 7 feet high, and has 3 doors, 6 feet 2 inches

by 2 feet 8 inches, and 2 windows, 4 feet by 2 feet 6 inches?

Solution: In the wallpaper example, it was found that there were 364 2/3

square feet after excluding the area in doors and windows.

Then Q-2Q X 364 2/3 ^ 0.2 x 364.667

100 100 0.2 X 3.64667,

which is about 0.73, or 3/4 gallon.

APPENDIX

To save work . . . Try this way of solving some simple problems.

Steps

1. Draw a horizontal division line.

2. Make entries above and below this line, depending on whether

the operation requires multiplication or division.

3. Cancel where possible. Cancellation often will eliminate much

of the work and speed up your calculations.

50
ExampI] ^ordwood is 40 feet long, 6 feet wide, and 5 feet high. How

much is the wood worth at $16 per cord? A cord occupies a space 8x4x4 feet.

Solution:

40 X 6 X 5

Cubic feet

in stack

8x4x4

Cubic feet

in one cord

Cords

X 16 $150

Value

per cord

Example #2: You and a friend plan a trip. You will use your car and he will pay

two-thirds of the cost of gasoline. If you average 15 miles per gallon of gasoline,

and its cost averages 30 cents per gallon, what will be your share of the gasoline

for a round trip totaling 1,200 miles?

Solution:

CQ

a

Co

<J

•iH

•4-»

U

>

CO

m

o

u

í)

as

i-i

O)

CJ

U

1,200

15

Gallons

x 0.30

Cost of gasoline

X 3

Your one-third

$8

51
Table 12.--Conversion factors—'

CommodityApples

Barley

Barley ñour

Beans, snap or wax

Calves

Cattle

Chickens

Com

Corn, shelled

Cotton

Cottonseed meal

Cottonseed oil

Dairy products

Butter

Cheese

Nonfat dry milk

Eggs

Flaxs eed

Hogs

Oats

P e anuts

Rye

Rye flour

Sheep and lambs

Soybeans

Soybean meal

Soybean oil

Tomatoes

Turkeys

Wheat

Wheat flour

1 pound dried

1 metric ton^/

100 pounds

1 case canned^/

1 pound live weight--

1 pound live weight--

1 pound live weight--

1 metric ton_2/

1 bushel (56 pounds)-

1 pound ginned1 pound1 pound1 pound

1 pound

1 pound

1 case

1 bushel

1 pound live weight1 metric ton^/

1 pound shelled

1 metric ton_2/

1 metric ton_2/

1 pound live weight1 metric ton^/

1 pound

1 pound

1 case canned^/

1 pound live weight1 metric ton_2/

100 pounds

Approximate equivalent

8 pounds fresh (beginning 1943)

45.93 bushels

4.59 bushels barley

0.010 tons fresh

0.555 pounds dressed weight

(since 1952)

0.549 pounds dressed weight

(since 1952)

0.72 pound ready-to-cook weight

39.368 bushels

2 bushels (70 pounds) of husked

corn

3.26 pounds seed cotton including

trash4/

2.10 pounds cottonseed

5.88 pounds cottonseed

21.1 pounds milk

10 pounds milk

11 pounds liquid skim milk

47 pounds

About 2 1/2 gallons oil

0.569 pounds dressed weight excluding lard (since 1952)

68.8944 bushels

11/2 bushels unshelled

39.368 bushels

2.23 bushels (beginning 1947)

0.477 pounds dressed weight

(since 1952)

36.744 bushels

1.28 pounds soybeans

5.45 pounds soybeans

0.027 ton fresh

0.80 pound ready-to-cook weight

36.7437 bushels

2.30 bushels wheat 5/

\l A more complete list may be found in the front pages of the annual publication

"Agricultural Statistics" of the U. S. Dept. of Agriculture. A table of weights and

measures also is shown.

21 A metric ton is equivalent to 2204.6 pounds.

3/ Case of 24 number 2 cans. 4/ Varies widely by method of harvest.

5/ This is equivalent to 4.51 bushels of wheat per barrel (196 pounds) of flour and

has been used in conversions beginning July 1, 1957.

52
Table 13.--Measurement equivalents and formulas

CONVERSION EQUIVALENTS

12 inches = 1 foot

3 feet = 1 yard

5,280 feet = 1 mile

16 1/2 feet = 1 rod

320 rods = 1 mile

144 square inches = 1 square foot

9 square feet = 1 square yard

1 inch = 2.54 centimeters

1 meter = 39.37 inches

1 square mile = 640 acres

60 miles per hour = 88 feet per second

1,728 cubic inches = 1 cubic foot

27 cubic feet = 1 cubic yard

231 cubic inches = 1 gallon (liquid

measure)

268.8 cubic inches = 1 gallon (dry

measure)

2,150.42 cubic inches = 1 bushel

1 cubic foot water weighs 62.4 pounds

1 cord = 128 cubic feet

31 1/2 gallons = 1 barrel

1 pint = 16 ounces

1 quart = 32 ounces

1 peck = 8 quarts

1 bushel = 4 pecks

MEASUREMENT FORMULAS

Circumference of circle = 3.1416 x diameter = =

Area of circle = .7854 x diameter x diameter

.Area of rectangle - length x width

Area of triangle = 0.5 x base x altitude

Area of curved surface of cylinder (like a silo) = 3.1416 x diameter x height

Volume of sphere = .5236 x diameter x diameter x diameter

Volume of cylinder = .7854 x height x diameter x diameter

Volume of pyramid = 1/3 x area of base x altitude

Volume of cone (like a stack of grain) = .2618 x height x diameter x diameter

53

* U.S. GOVERNMENT PRINTING OFFICE : 196Z 0 — 636983
Çrowth Through Agricultural Progress

i