22 February 2002

from: Dennis Couzin

to: ASTM-Flash

re: **2, 3, many Blondel-Reys**

This note probes Blondel & Rey\'s 1911 work, as portrayed in Douglas\'
1957 paper[^1]. The men\'s names are merely labels for the equations
which are the proper objects of this study.

![](media/image1.wmf){width="5.125694444444444in"
height="3.2604166666666665in"}

FIGURE 1

A pulse of light is given by its intensity function I(t).

For any two times t~1~, t~2~ during the pulse we can define a quotient
Q:

> $Q \equiv \frac{\int_{t = t_{1}}^{t_{2}}{I(t)\ \text{dt}}}{t_{2} - t_{1} + a}$
> (1)

a is a time constant.

If I(t) is a rectangle pulse of intensity I starting at t~1~ and ending
at t~2~ then Q gives the Blondel-Rey effective intensity (I~e~) of the
pulse as in equation (2)

$I_{e} = \frac{(t_{2} - t_{1})\ I}{t_{2} - t_{1} + a}$ (2)

Blondel & Rey wanted Q to give the effective intensity of a general
pulse. All that is needed is to specify t~1~ and t~2~.

Obviously t~1~ and t~2~ can\'t just be the pulse\'s starting and ending
times because the general pulse could start very slow or drag on. Using
earliest or latest times enlarges the denominator, reducing Q,
independent of the flash visibility.

According to Douglas, Blondel & Rey wanted t~1~ and t~2~ chosen so as to
maximize Q. Why did they want this? Did they think of Q as measuring
some effect of I(t) between t~1~ and t~2~, and think of visual detection
as a *maximizing scan* over all these effects? I find this idea nutty. A
neural process can scan through a time range, but it can\'t conduct a
two dimensional scan over time time space. Nevertheless it is called a
principle here, in the hope that someone will explain why it isn\'t
absurd.

**MaxScan principle:** Effective intensity of flash I(t) is the maximum,
over all t~1~,t~2~ pairs, of the value of some operation applied to I(t)
between limits t~1~ and t~2~.

An example of an operation is Blondel & Rey\'s Q(I,t~1~,t~2~).

Formally:

> $I_{e} = Q_{\text{max}} = \underset{t_{1},t_{2}}{\text{Max}}\left\{ \frac{\int_{t = t_{1}}^{t_{2}}{I(t)\ \text{dt}}}{t_{2} - t_{1} + a} \right\}$
> (3)

Despite its odd origin, equation (3) is mathematically impeccable. It
works for every flash I(t), regardless of complexity. Equation (3) gives
no hint which t~1~ and t~2~ maximize Q. Try\'em all; there must be a
maximum. According to the MaxScan principle, it shouldn\'t matter
whether more than one t~1~,t~2~ pair produces this maximum.

**Blondel-Rey Equation**[^2]

Blondel & Rey\'s famous equation does give a hint how to find t~1~ and
t~­2~:

$I_{e} = \frac{\int_{t = t_{1}}^{t_{2}}{I(t)\ \text{dt}}}{t_{2} - t_{1} + a}$
t~1~,t~2~ such that I(t~1~) = I(t~2~) = I~e~ (4)

Equation (4) says to choose t~1~ and t~2~ so that the resulting Q is
equal to both I(t~1~) and I(t~2~).

Douglas more or less proves that Blondel & Rey\'s equation (4) achieves
their desideratum (3) for simple flashes. He is unclear whether Blondel
& Rey knew this. Blondel & Rey might have arrived at (4) intuitively,
independently of (3). I can\'t see their intuitions behind (4). They
might have found (4) by fiddling with examples of (3).

Equation (4), the \"Blondel-Rey equation\", has twisted minds for 90
years. Here is a proof that the equation has a solution.

**Existence proof**

Try out two different I~e~ for the left hand side of equation (4),
always choosing t~1~ and t~2~ as far apart as possible, with I(t~1~) =
I(t~2~) = I~e~.

First try 0 for I~e~. t~1~ is at the beginning of the flash and t~2~ at
the end. This makes the right hand side of equation (4) positive, that
is, greater than the left hand side.

Next try I~max~ for I~e~. For a flash that reaches I~max~ at a single
peak, I(t~1~) = I(t~2~) = I~max~ forces t~1~ = t~2~, which makes the
right hand side of equation (4) equal 0, that is, less than the left
hand side. For a flash that reaches I~max~ at more than one t, the right
hand side numerator is equal or less than (t~2~-t~1~)I~max~, while the
denominator is greater than (t~2~-t~1~). Therefore the right hand side
is less than I~max~, that is, less than the left hand side.

With this method of choosing t~1~ and t~2~, the right hand side of
equation (4) is a continuous function of I~e~. Therefore our finding
that I~e~ = 0 makes the right hand side greater than I~e~, and I~e~ =
I~max~ makes the right hand side less than I~e~, proves that some value
of I~e~ between 0 and I~max~ makes the right hand side equal to I~e~,
i.e., solves the equation. QED

It can also be proved that solution to the Blondel-Rey equation is
unique.

**Uniqueness proof**

Equation (4) can be rewritten:

$\text{aI}_{e} = \int_{t = t_{1}}^{t_{2}}{\left( I(t) - I_{e} \right)\ \text{dt}}$
t~1~, t~2~ such that I(t~1~) =I(t~2~) =I~e~ (5)

Suppose the I~e~ in (5) is the larger of two solutions. The smaller I~e~
makes the left hand side of (5) smaller. For a single peaked flash,
smaller I~e~ requires t~1~ and t~2~ to be farther apart. This makes the
right hand side of (5) larger. Equation (5) then fails. QED

Equation (4) has a unique solution and equation (3) has a unique
solution. This is where Douglas starts his proof that the equations have
the same solution. His proof misses one of three cases, but it is
basically correct for the flash with a single peak. Here is a more
rigorous proof that (4) is equivalent to (3), which allows extension to
multi-peaked flashes:

**Maximality proof**

Suppose t~1~ and t~2~ solve equation (3), and t~1~ t~2~. It must first
be shown that I(t) is sloping upwards at t~1~ and downwards at t~2~.
This is easy for a single peaked flash.

Suppose t~1~ and t~2~ are on the same slope. Suppose I(t) is sloping
upwards at times t~1~ and t~2~. I(t) is still sloping upwards at
slightly later times t~1~+ and t~2~+. The two later times produce a
larger numerator in the Q quotient, while leaving the denominator
unchanged, so the new Q exceeds Q~max~.

Similarly if I(t) is sloping downward at times t~1~ and t~2~.

It is less easy shown for a multi-peaked flash, so that part of the
proof is postponed. See Straddling proof, below.

Now suppose I(t~1~) I(t~2~).

Case I, Q~max~ I(t~1~). Try a slightly later initial time t~1~+ where
I(t~1~)I(t~1~+) Q~max~. This decreases the numerator of the Q quotient
by less than Q~max~ while decreasing the denominator by , so the new
quotient exceeds Q~max~.

Case II, Q~max~ I(t~1~). Try a slightly earlier initial time t~1~- where
Q~max~ I(t~1~-). This increases the numerator of the Q quotient by more
than Q~max~, while increasing the denominator by , so the quotient
exceeds Q~max~.

Case III, Q~max~ = I(t~1~). If I(t~1~) I(t~2~), try a slightly earlier
final time t~2~- where\
Q~max~ = I(t~1~) I(t~2~-) I(t~2~). This decreases the numerator of the Q
quotient by less than Q~max~ while decreasing the denominator by , so
the new quotient exceeds Q~max~.

Therefore Q~max~ = I(t~1~) = I(t~2~) as in equation (4). QED

Thus FIGURE 1 with I(t~1~) I(t~2~) can\'t show a solution to equation
(3). Solutions must look like FIGURE 2, which also illustrates equation
(4).

![](media/image2.wmf){width="5.125694444444444in"
height="3.2604166666666665in"}

FIGURE 2

Did Blondel & Rey countenance multi-peaked flashes? When these flashes
are perceived as single, the concept of effective intensity must apply,
and Blondel & Rey\'s MaxScan principle should apply. Then their equation
(3) gives the value of I~e~.

FIGURE 3 shows how t~1~ and t~2~ might be chosen to solve equation (3).
Equation (3) never fails, but is difficult to apply because all
t~1~,t~2~ pairs must be considered. Can the Blondel-Rey equation (4) be
applied to a multi-peak flash? In the proof above, the first (Existence)
part works for a multi-peak flash. Equation (4) will have a

![](media/image3.wmf){width="5.125694444444444in"
height="3.2604166666666665in"}

FIGURE 3

solution for any flash. It is the second (Uniqueness) part which
explicitly requires that the flash have one peak. Equation (4) can have
multiple solutions for multi-peak flashes. FIGURE 4 shows an example
where there are three solutions. The lowest solution (t~1~,t~2~)
encompasses just the first pulse. The middle solution (t~3~,t~4~)
encompasses both pulses. The highest solution (t~5~,t~6~) encompasses
just the second pulse.

![](media/image4.wmf){width="5.125694444444444in"
height="3.2604166666666665in"}

FIGURE 4

For multi-peak flashes, maximality means that the highest solution of
equation (4) equals the solution of equation (3). For the third
(Maximality) part of the proof to apply to a multi-peak flash, it
remains to be proved that for t~1~,t~2~ solving equation (3), I(t)
slopes upwards at t~1~ and downwards at t~2~. That is, t~1~ and t~2~
straddle one or more peaks. For example, solutions to equation (3)
cannot look like FIGURE 5.

![](media/image5.wmf){width="5.375in" height="3.2715277777777776in"}

FIGURE 5

**Straddling proof**

Suppose the solution of equation (3), Q~max~, is achieved using t~1~
t~2~.

Can I(t) have downward slope at t~1~?

Case I: I(t~1~) Q~max~. Reduce t~1~ slightly to t~1~- so I(t~1~-)
Q~max~. The numerator of the Q quotient increases by more than Q~max~.
The denominator increases by . So the new quotient exceeds Q~max~.

Case II: I(t~1~) Q~max~. Increase t~1~ slightly to t~1~+ so still
I(t~1~+) Q~max~. The numerator of the Q quotient decreases by less than
Q~max~. The denominator decreases by . So the new quotient exceeds
Q~max~.

Can I(t) have upward slope at t~2~?

A mirror image argument to the last shows it can\'t. QED

With Straddling proved, the Maximality proof now handles the multi-peak
case. The solution to equation (3) also satisfies equation (4). The
maximal solution to equation (4) is the solution to equation (3).
Equation (4) allows fairly practical numerical solution. Different
heights are tried as I~e~ solutions. For each height, the intersections
with I(t) give several t\'s. Q is calculated for all t~1~ t~2~ pairs for
which I(t) is rising at t~1~ and falling at t~2~. The convergence to
I(t~1~) = I(t~2~) = I~e~ is less simple than with single peaked flashes.
If there are N peaks the search can proceed through as many as N(N+1)/2
paths.

**Blondel-Rey-Douglas**

Douglas (1957) proposes an alternative equation (6) to the Blondel-Rey
equation (4) specifically for use with multi-peaked flashes.

$$I_{e,D} = \frac{\int_{t = t_{1}}^{t_{2}}{I(t) \cdot \left( \frac{1 + \text{sign}\ \left( I(t) - I_{e,D} \right)}{2} \right)\ \text{dt}}}{t_{2} - t_{1} + a}$$

> t~1~, t~2~ such that I(t~1~) = I(t~2~) = I~e,D~ (6)

Instead of Blondel-Rey\'s I~e~, this equation yields a new effective
intensity called here I~e,D~. Douglas didn\'t write his equation as (6)
is written. He wrote a sum of separate integrals in the numerator and
included a diagram showing how the integration skipped the times between
t~1~ and t~2~ where I(t) fell below I~e,D~[^3]. The term in big
parentheses in the numerator of equation (6) becomes 1 for all t where
I(t) I~e,D~ and becomes 0 for all t where I(t) I~e,D~. The signum
function is defined:

sign(x) = -1 for x 0

sign(0) = 0

sign(x) = +1 for x 0.

The definition at 0 is incidental. Equation (6) is a practical
equivalent to the equation Douglas wrote.

Douglas gave incomplete rules for application of equation (6). I believe
he meant it to apply when gaps between pulses were between 10 ms and 100
ms, but he wrote that it should be applied when gaps are less than 10
ms. I do not understand Douglas\' thinking which led to equation (6).

We cannot regard Douglas\' quotient as a new quotient Q\' replacing the
Q quotient of Blondel-Rey. We cannot write a straightforward
maximization equation standing to equation (6) as equation (3) stands to
equation (4). I~e~ figures implicitly in both sides of equation (4),
since the choice of t~1~ and t~2~ depends on I~e~. In equation (3), t~1~
and t~2~ are free, and are chosen to maximize Q. However, I~e~ figures
explicitly in both sides of equation (6). Choosing a pair t~1~,t~2~ is
not enough to allow calculation of Douglas\' quotient.

Equation (6) is so sick that it can lack a solution.

![](media/image6.wmf){width="5.386111111111111in"
height="3.0347222222222223in"}

FIGURE 6

FIGURE 6 shows an example where there is no I~e,D~. The time axis is in
seconds.

> I(t) = 1 cd from t = 0 to t=0.05.
>
> I(t) = 0.25 cd from t = 0.05 to t = 0.25.
>
> I(t) = 1 cd from t= 0.25 to t = 0.30.
>
> Time constant a = 0.2 sec.

The only possible solutions can be with

> interval A: t~1~ = 0, t~2~ = 0.05
>
> interval B: t~1~ = 0, t~2~ = 0.3
>
> interval C: t~1~ = 0.25, t~2~ = 0.3

Interval A doesn\'t work because the initial pulses can muster only
I~e,D~ =
$\frac{1 \times 0\text{.}\text{05}}{0\text{.}\text{05} + 0\text{.}2} = 0\text{.}2$
cd, but I~e,D~ must equal at least 0.25 cd in order to have I(0.05) =
I~e,D~.

Similarly interval C doesn\'t work.

For interval B, equation (6) says to exclude from the integral the time
where I(t) falls below I~e,D~. If the whole time interval is included,
then:

I~e,D~ =
$\frac{1\  \times \ 0\text{.}\text{05} + 0\text{.}\text{25}\  \times \ 0\text{.}2 + 1\  \times \ 0\text{.}\text{05}}{0\text{.}3 + 0\text{.}2}$
= 0.3 cd

But this means the 0.25 cd time in the center must be excluded. When
this is excluded:

I~e,D~ =
$\frac{1\  \times \ 0\text{.}\text{05} + 1\  \times \ 0\text{.}\text{05}}{0\text{.}3 + 0\text{.}2}$
= 0.2 cd

But this means the 0.25 cd time in the center must be included.

Contradiction! Equation (6) has no solution for the simple flash of
FIGURE 6.

The trouble with equation (6) is in its mathematical form. The
contradictory example of FIGURE 6 does not rely on the perfectly square
flash shape. Also, similar examples can be constructed for shorter
times, as required by Douglas.

Douglas\' modification of the Blondel-Rey equation is intuitively
mysterious and mathematically incoherent. It was not a serious
contribution to the science of flashing lights, and it will be omitted
from the next version of my flashcalculator spreadsheet.

Blondel & Rey\'s equation (3), or equivalently the maximum solution to
their equation (4), will be included in the next version of the
spreadsheet.

[^1]: Charles A. Douglas, Computation of the Effective Intensity Of
    Flashing Lights, Illumination Engineering, December 1957, pp.
    641-646.

[^2]: Douglas, p. 641: \"An equation of this form was originally
    suggested by Blondel and Rey \[1911\], but has rarely been used.\"

[^3]: Douglas, p. 644 equation (12), p. 645 Figure 5.