- xXIHAYD0IXx
- |
- Intrepid Legendary Member
This is the way the world ends,
This is the way the world ends,
This is the way the world ends;
Not with a bang but a whimper.
It depends how BA you want your Spartans to seem. Minimum distance to deploy a chute and still have enough elevation to deploy a reserve should the primary fail is 1500ft (~460m) AGL but it only takes 300ft (~90m) AGL for fulpounds to kilogramsl chute deployment. Keep in mind that lowest allowed altitude for a combat jump is 800ft (~240m) AGL.Note that most Halo (High Altitude Low Opening) jumps occur between 2000 and 3000ft (~600-910m) AGL. If your Spartans are only weighing about 1000lbs (assuming light armor) I would go with a chute deployment of between 1000 and 2000ft (~300-600m) AGL. It'd make them seem awesome without making them seem suicidal. But it is, of course, your decision to make.
Your terminal velocity question is rather more difficult. I have to make a number of assumptions:
Assumption 1: Each Spartan weighs 200lbs with the suit weighing 800lbs. This makes the average density roughly 5000kg/m^3.
Assumption 2: The Spartans are 0.5m (~20in) broad and 2m (~7ft) tall.
Assumption 3: The Spartans are diving at a steep 60 degree angle, yielding a cross-sectional area of about 0.5m^2.
Assumption 4: The drag coefficient of a Spartan is 1.0 (don't ask for units).
Assumption 5: Reach is a 1G planet.
Each of these assumptions are within reason, although they are still assumptions. Unless your reader is a physicist, they should not be able to tell the difference.
In any case, your answer is: 108 m/s for their terminal velocity. That said, they would accelerate from about 650km until around 100km, assuming Earth-like conditions. So terminal velocity does not matter until hitting around 100km because of a lack of atmosphere. Sadly, I was too lazy to bust out the calculus, so I made an estimate for the density of air. vT would be higher at greater elevations. It should still be a decent estimate provided you do not go above 100km in your calculations.
This is all using the formula vT=sqrt[(2mg)/(pACd) where m is mass, g is force of gravity, p is density of fluid, A is applicable area, and Cd is drag coefficient. I hope this helps somewhat. Keep in mind, the number could be totally wrong, so I take no responsibility if some PhD reads your work and slams you for it.
EDIT- Given terminal velocity of a skydiving human is about 55m/s. Considering that sqrt(5)=2.2, I am somewhat more confident in the veracity of my answer.
EDIT 2- I felt bad about leaving your question vauge. vT at sea level=108m/s. vT at 50km=3770m/s. vT at 100km=119,350m/s. If your spartans are reaching a thirtiet of the speed of light, you have an issue. So start your calculations at 50km up. I'll be back with a calculus answer ASAP.
[Edited on 08.14.2012 8:20 PM PDT]