# Project Longshot/Appendix
## APPENDIXAPPENDIX
### Delta-V CalculationsDelta-V Calculations
#### Assumptions:Assumptions:

   * space station orbit altitude = 300 km
   * space station orbit inclination = 28.5 deg
   * obliquity = 23.5 deg
   * 1 AU = 149.6 x 10$^6$ km
   * Earth radius = 6378 km
   * f$_{earth}$ = GM$_e$ = 398,601.2 km$^2$/sec$^2$
   * f$_{sun}$ = GM$_e$ = 1.3271544 x 10$^{11}$ km$^2$/sec$^2$
   * f$_{Jupiter}$ = GM$_e$ = 1.268 x 10$^8$ km$^2$/sec$^2$
    (1) Velocity of probe about the earth:
       V$_p$ = (GMe/Rcircular).5 = (398,601.2/6678)$^{.5}$ = 7.7258 km/sec
    (2) Amount of plane change:
       i = 61x − (28.5x + 23.5x) = 9x
    (3) Delta-V for 9x plane change:
       Delta-V$_{pc}$ = 2 x V$_p$ x sin(i/2) = 1.2123 km/sec
    (4) Earth escape velocity:
       V$_{e−escape}$ = (2GMe/R)$^{.5}$ = 10.9260 km/sec
    (5) Delta-V to escape earth:
       Delta-V$_{e−esc}$ = V$_{e−esc}$ - V$_p$ = 3.2002 km/sec
    (6) Velocity of probe around sun:
       V$_{p−sun}$ = (f$_{sun}$/1 AU).5 = 29.7849 km/sec
    (7) Solar system escape velocity:
       V$_{sun−escape}$ = (2f$_{sun}$/1 AU)$^{.5}$ = 42.1221 km/sec
    (8) Delta-V to escape the solar system:
       Delta-V$_{sun−esc}$ = V$_{sun−esc}$ - V$_{p−sun}$ = 12.4273 km/sec
    (9) Total Delta-V required to escape solar system:
       Delta-V$_{total}$ = 16.8398 km/sec ​
Rough calculations for Delta-V's for orbit changes to take the probe out to Jupiter (to take on fuel) are made using Hohmann transfer equations as an approximation (actual orbits would be patched-conics). The first six calculations are the same for this analysis:
    (1) Energy of Hohmann transfer orbit:
       E$_t$ = −f$_{sun}$/6.2 AU = −143.0894 km$^2$/sec$^2$
    (2) Perigee velocity of transfer orbit:
       V$_1$ = [2(f$_{sun}$/1 AU + E$_t$)]$^{.5}$ = 38.6171 km/sec
    (3) Delta-V to enter transfer orbit:
       Delta-V$_1$ = V$_1$ − V$_{p−sun}$ = 8.8223 km/sec
    (4) Apogee velocity of transfer orbit:
       V$_2$ = [2(f$_{sun}$/5.2 AU + E$_t$)] = 7.4157 km/sec
    (5) Velocity required to orbit Jupiter at an altitude of 500,000 km:
       V$_3$ = (f$_{Jupiter}$/571370)$^{.5}$ = 14.8971 km/sec
    (6) Delta-V to inject into orbit about Jupiter:
       V$_3$ − V$_2$ = 7.4814 km/sec
    (7) Jupiter escape velocity:
       V$_{3−escape}$ = (2f$_{Jupiter}$/571370)$^{.5}$ = 21.0676 km/sec
    (8) Delta-V to escape Jupiter:
       Delta-V$_{3−esc}$ = V$_{J−esc}$ − V$_J$ = 6.1705 km/sec
    (9) Velocity of probe about sun at Jupiter distance:
       V$_{p−sun−J}$ = (f$_{sun}$/5.2 AU)$^{.5}$ = 13.0601 km/sec
    (10) Velocity to escape sun at Jupiter distance:
       V$_{sun−esc−J}$ = (2f$_{sun}$/5.2 AU)$^{.5}$ = 18.4697 km/sec
    (11) Delta-V to escape sun at Jupiter distance:
       V$_{sun−esc−J}$ - V$_{p−sun−J}$ = 5.4087 km/sec
    (12) Total Delta-V for Jupiter analysis Delta-V
       V$_{total}$ = 32.2954 km/sec
This is nearly twice the Delta-V required for the first case. This fact, along with the difficulties involved in mining the atmosphere of Jupiter, getting the fuel to a million kilometer orbit around the planet, etc., has made it obvious that obtaining fuel from Jupiter is not feasible.
Acceleration and Velocity Profiles
1. Accelerations. The accelerations for the mission were found using the equation F=ma. Since it is a constant thrust, the acceleration will increase at a constant rate. The acceleration was found by dividing the thrust by the mass left at the time. This gives the acceleration chart on page.
| Probe Wt. | No. SRB's Shuttle Size | Delta V | S/C TOTAL Wt. |
| mt | | km/s | mt |
| 1409.695 | .1 | .205 | 1469.3 |
| 1409.695 | .2 | .401 | 1528.905 |
| 1409.695 | .3 | .589 | 1588.51 |
| 1409.695 | .4 | .768 | 1648.115 |
| 1409.695 | .5 | .94 | 1707.72 |
| 1409.695 | .6 | 1.104 | 1767.325 |
| 1409.695 | .7 | 1.263 | 1826.93 |
| 1409.695 | .8 | 1.415 | 1886.535 |
| 1409.695 | .9 | 1.562 | 1946.14 |
| 1409.695 | 1 | 1.703 | 2005.745 |
Plane change of 9 degrees Delta V required is 1.2123 km/s
| Probe Wt. | No. SRB's Shuttle Size | Delta V | S/C TOTAL Wt. |
| mt | | km/s | mt |
| 694.435 | 1 | 2.914 | 1280.485 |
| 694.435 | 1.1 | 3.112 | 1350.09 |
| 694.435 | 1.2 | 3.298 | 1409.695 |
| 694.435 | 1.3 | 3.475 | 1469.3 |
| 694.435 | 1.4 | 3.643 | 1528.905 |
| 694.435 | 1.5 | 3.802 | 1588.51 |
Escaping Earth Delta V required is 3.2002 km/s
| Probe Wt. | No. SRB's Shuttle Size | Delta V | S/C TOTAL Wt. |
| mt | | km/s | mt |
| 396.41 | .1 | .689 | 456.015 |
| 396.41 | .2 | 1.28 | 515.62 |
| 396.41 | .3 | 1.795 | 575.225 |
| 396.41 | .4 | 2.249 | 634.83 |
| 396.41 | .5 | 2.653 | 694.435 |
Aiding in escaping solar system Delta V required is at least 2.5 km/s
Laser-Pumped Light Sail
    Assumptions:
       1) 100% of laser is focused onto the sail continuously for one year.
       2) the solar energy to pressure ratio holds true for the specific laser used.
    At 1 A.U.:
       1353 watts/square meter from the sun
             and
       4.6E-6 N/square meter from the sun
    therefore,
       2.94E8 watts/N.
    Payload Mass = 30,000 kg
    Delta V required = 13500 km/sec
    Time interval = 31,472,262 sec (1 year)
    Therefore,
       Required acceleration = .429 m/sec squared
             and
       The force required = 1268.8 N
    So that with a laser operating at 100% efficiency
       Required power = 3.78E12 watts
EXTREME HIGH TEMPERATURE EXPANSION OF GAS
"Ideal" Rocket Nozzle: Perfect Gas
          Steady Flow - no shock, friction or heat losses
          One Dimensional Flow
          Frozen Chemical Equilibrium
    V$_e$ (exit velocity) = L E
    L = Limiting Gas Velocity
    E = Pressure Expansion Ratio
       $E={\sqrt {1-{\frac {P_{e}}{P_{t}}}^{\frac {\gamma -1}{\gamma }}}}\quad \quad (P_{e}=0)$
             $L={\sqrt {\frac {2\gamma {\overline {R}}T_{t}}{(gamma-1)M}}}$
    $\gamma _{(monoatomic)}=1.67$
    ${\overline {R}}=8.31441$
    $M_{H_{1}}=1.00794$
    $T_{t_{critical\,deuterium\,fusion}}=3.5x10^{8}\,\,\,K$
    $L_{\mathrm {max} }=379{\frac {km}{sec}}$
    $I_{sp}={\frac {V_{e}}{g}}=39000\,\mathrm {sec}$
Fusion Reaction
    $\mathrm {He_{3}+H_{2}\rightharpoonup He_{4}+H_{1}}$
    at reaction temperature:
    ${\begin{alignedat}{2}\mathrm {He_{4}} &=\mathrm {\alpha _{4}^{2+}+2e^{-}} \\\mathrm {H_{1}} &=\mathrm {p_{1}^{+}+e^{-}} \\\end{alignedat}}$
the resulting charged particles can be magnetically funneled and used to charge induction coils upon exiting.
Pellet Size vs. Frequency
an engineering analysis must be done on the proposed engine to determine pellet size and pulse frequency.
PROCESSING BLOCK DIAGRAM
Sizing of the Fuel Tanks
1. Tank Volume
The storage density for the fuel is .0708 tons per cubic meter. Therefore, the total volume needed is:
| Vol | = (264.276 mt)*(1/.0708 mt/m^3) |
| | = 3732.71 m^3 |
The radius of the fuel tanks is set at 2.5 meters and the length is then computed.
| length | = (Vol/6)/(pi*r^2) |
| | = (3732.71 m^3/6) / (pi*(2.5)^2) | = 31.68 meters |
The total volume is divided by six, because six fuel tanks are used.
2. Tank Thickness
A. Interstellar cruise phase
Newton's equation, F=ma, is used to find the forces on the fuel tanks during flight. The maximum force will occur when the acceleration is a maximum. This is when the mass is a minimum for a constant thrust problem. In our case, the minimum mass is:
| Mmin | = 2 fuel tanks + 6 straps + center section + engine + payload |
| | = 89.345 metric tons |
Therefore,
| Amax | = Thrust/m |
| | = 1.838e3 N / 89.345e3 kg |
| | = .021 m/s^2 |
The maximum force possible on the tanks could then be:
| Fmax | = (mass of fuel in tanks) * a |
| | = 44.05e3 kg * .021 m/s^2 |
| | = 924.97 N |
A safety factor of 5 is now applied to find the design load.
| Fdes | = 5 * F |
| | = 4.62 kN |
Next, the tank area that the force is acting on is found. ​
| Area | = pi * r^2 |
| | = pi * 2.5^2 |
| | = 19.635 m^27 |
So, the pressure will be:
| p | =F/A |
| | = 4.62 kN / 19.635 m^2 |
| | = 235.54 N/m^2 |
The thickness of the tanks is now determined.
t = (P * r) / (2 * yield stress)
If we assume that we will use A1 1100-0, the yield stress is 3.45e7 N/m^2. This gives a thickness of:
| t | = (235.54 N/m^2*2.5 m)/(2*3.45e7 N/m^2) |
| | = 8.53e-6 m or .00853 mm |
Therefore, the fuel tanks can be made from 2 mm A1 1100-0 sheets.
B. Orbital Maneuvers
Since the upper stages will require advanced upper stages, the forces on the spacecraft are not known. Through the use of staging, multiple burns per maneuver, and the use of a stronger aluminum alloy, the thickness of the fuel tanks can be kept at 2 mm.
WEIGHT TABLE
| Item | Components | Weight Metric Tons |
| Engine |
| | Chamber | 4.35 |
| | Igniter | 17.543 |
| | Field Coil | 10.245 |
| | Total | 32.138 |
| Fuel | Total | 264.276 |
| Structure |
| | Fuel Tanks | 55.038 |
| | Straps | 9.151 |
| | Center Section | 1.737 |
| | Other Structure | 4.074 |
| | Total | 70.00 |
| Payload |
| | Reactor | 10.00 |
| | Instruments | 3.00 |
| | Lasers | 2.00 |
| | Misc. (shielding, etc...) | 15.00 |
| | Total | 30.00 |
| | Overall Total | 396.414 |