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How do I know how many parachutes I need?

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So far I've been guessing on how many parachutes I need to recover a ship intact, and I find that I often come down hard enough to destroy various pieces of my craft. Can I solve this with math?

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Calculating drag now requires a computer simulation to get accurate values. With that said, I have been using the calculator Invader Jim posted to get a ballpark idea of the number of parachutes to use.

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Can I solve this with math?

Yes, but with the new aero it is rather involved and requires inputs that are not easily obtained. Your are probably better off determining it experimentally.

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My rule is 2 on the cockpit, and 1 per every 2nd module. Of course I'm talking about a 3-5 module/part ship which is quite small. Jettison your tanks and you usually don't have to worry about weight too much; 3 parachutes is generally enough to land at 5m/s.

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This is a multidimensional analysis involving (in order):


Atmospheric Density

Local Gravity


Terminal Speed

Vessel Weight

Free Parameter

Total Drag (Parachutes)

In general a full understanding of "how will the craft behave under parachute(s)" is a multi-dimensional solution space. But mission requirements can fix several of these to reduce the dimensionality of the solution space to something more useful.

(A) The first to be fixed is likely the atmospheric density which is characteristic of the landing environment. The density is strictly a function of altitude per body. Every atmosphere of every body is simply density in character there exists an altitude on every body that is representative of an altitude on Kerbin. Provided that altitude is accessible (greater than 0m) one could happily test parachutes for a Duna surface mission on Kerbin at some appropriate altitude. More atmosphere means a slower craft in a linear way I believe with stock.

(G) The second to be fixed again by environment is local gravity to which surface gravity is often a perfectly adequate. Gravity strength is multiplicative with vessel mass to determine weight. More weight means a faster craft, again linear.

(S) The vessel by construction will have some maximum impact speed to safely land. To be the best of my knowledge this is the impact strength of the part which comes into contact with the surface or the strength of the joints which hold together the parts. Landing legs which survive 100 m/s impact are of little use if the forces rip apart the craft. Stronger craft require less parachutes which can save weight but often that robustness is heavy in its own right.

(M) Vessel mass is again due to design. The more massive the craft the faster it will fall.

(P) Drag from parachutes counteracts the falling speed. Each parachute will provide drag proportional to the speed.


Assuming the landing happens at terminal velocity, the relationship is very straightforward. Downward force and upward force are equal. Downward force is mass (M) and gravity (G) and upward force is the atmosphere density (A) times the parachute factor (P) times speed (S)*. *squared.

M x G = P x S^2 x A

Given M, G, S, and A are known, P is determined:

P = M x G / S^2 x A

All it would take is to look up what "P" values various parachutes are, plugging in some numbers and rounding up as a safety factor. The first step is to calibrate this equation to some meaningful values for known parachute parts and units of M G S and A. Without regard for the pleasantness of the number we might define 1 "P" parachute factor to maintain a 1 ton mass inside 1 m/s^2 of gravitational acceleration to a terminal velocity of 1 m/s in 1 MT/m^2 of density. I've tried to use the most KSP-native units here. I could just as easily defined a 1kg mass under Kerbin's surface gravity to a terminal velocity of 6m/s at Kerbin's surface at Kerbin's surface density. The former is more elegant and general while the latter might be most useful in common cases. Both tables of "P" values for various parachutes would be the same except for a multiplication constant.


Assuming that pressure is indicative of density (large changes of temperature and molecular composition invalidate this).

So a 920 kg craft:

under 9.70m/s^2 gravity is going 6.09m/s in an atmosphere of 58.75. (4.1)

under 9.75m/s^2 gravity is going 5.84m/s in an atmosphere of 76.45. (3.44)

under 9.78m/s^2 gravity is going 5.50m/s in an atmosphere of 86.74. (3.43)

under 9.81m/s^2 gravity is going 5.10m/s in an atmosphere of 99.66. (3.48)

Throwing out the first one and averaging the rest I get a "P" factor for a Mk16 parachute of 3.61. If it pleases you you can multiply this number by 1000 and input the craft mass in kg instead of metric tons and it is equivalent. Alternatively one could express this in atmospheres which would have an effect of moving the decimal nearly 2 places to the right. Doing both moves the decimal 5 places.


A fine theory but does it stand up to experiment? To Duna! Our familiar 920kg craft just as before reentering. What will our speed be? It should be the square root of MG/PA.

M 920 kg

G 2.86 m/s^2

P 3.61

A 11.97 kPa

7.8 m/s! Turns out it was closer to 10.8. Seemed pretty reliable on Kerbin but under reported on Duna. Was my math bad? Certainly possible. It's also possible the pressure is not a good indicator of density when switching from Kerbin to Duna. As far as I know Duna should be both colder and of a denser molecule compared to Kerbin which should improve parachute performance.

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When using a reentry vehicle that's dissimilar to what I've done before, I usually take just that part of it and put it on top of a SRB that's been tweaked to have just 2-3 seconds worth of fuel at a high TWR. Basically enough to get it up a few thousand meters. Launch it and stage the chutes and decoupler when it starts going down again. This also will verify if it's going to fly with the heat shield forward or not.

Sure, you can math it out - but sometimes it's easier to make the situation easily testable.

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By mass, the most efficient number of parachutes to have is.... one.

Plus a handfull of seperatrons firing at the very last 10m before landing.

Also consider investing is nice high-impact girders as landing legs.

Even a single mk16 parachute will slow a 50-ton ship to 25m/s on Kerbin at KSC.

So take your 50-ton hulk, add ONE parachute, and add 4 short girders.

The end solution is much lighter than the 12-16 parachutes you would have needed.

(not elegant at all, but so what)

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I'll test my parachutes when I test my whole rocket.

Keep testing until my rocket can make a fairly efficient gravity turn and circulize. Then Ill get to an apoapsis that's barely in kerbin soi, lower periapsis to about 20 km. Go through my reentry and parachute procedure.

This tests my gravity turn, ability to change an orbit by about 60-80 Mm in one burn, worst case scenario reentry and landing.

My actual reentries are never in that much risk since I'll lower periapsis to 30km if I'm going slow and 50km or so with a powered deceleration if I'm going fast, and plus I'll have less fuel weight and thus will slow down a lot faster.

If I'm just testing a craft that will stay in kerbin soi then I'll go straight up to save real world time

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