In earlier versions of Kerbal Space Program, even in Career Mode, parts cost and recovery was never an issue to be concerned about. So a typical space mission might only return to Kerbin with the command pod and perhaps some scientific equipment. This would typically just need one or two parachutes - rarely more, and you could usually just make a "gut instinct" judgement call to decide when to throw on one or two more.

Version 0.24, "First Contract", changed this significantly. Now, vehicle components have monetary value and that value can at least partially be refunded by returning the component to Kerbin safely. This doesn't just go for the stage containing the command pod, either. If you can get lower stages landed intact, you can recover those costs as well. This new responsibility to keep whatever parts we can greatly increases the complexity of spacecraft design in terms of the number and placement of parachutes.

After a bit of trial and error, I think I've more or less gotten a handle on placement. For small chunks that are good at landing vertically (and staying that way), it's no big deal - just use radial symmetry and put 'em wherever. Larger pieces need all of their parachutes on one side, and evenly distributed from the center of mass - otherwise, when the tail lands first and the 'chutes cut automatically (Why?), the rest of the component will free-fall to the ground and likely be destroyed on impact.

The problem I'm having though is figuring out how many parachutes are appropriate to safely land a given mass. A couple times, I've put on too few and some (or all) parts ended up destroyed when they touched down too fast. So what I usually wind up doing now is putting on too many parachutes, which causes the final descent to be unnecessarily and inexorably slow.

So, given the following known values:

  • Total Mass of components to be safely returned to Kerbin
  • Impact Tolerance of weakest component
  • Mass and Drag of each parachute model

Are there any easy ways to calculate how many parachutes will be appropriate to safely return the components to Kerbin without slowing them down much more than is absolutely necessary?

I'm sure there's a certain amount of complexity to the "right way" of doing it (and I'd definitely be interested in seeing that), but I'm really looking for something that's more easily done by those of us who aren't rocket scientists outside of KSP. A general rule along the lines of "You need x drag per unit of mass to reduce the impact speed by 1 m/s" or "Use x drag per unit of mass to reduce the impact speed to y m/s and z drag per unit of mass per 1 m/s after" would be preferable, if the problem can be indeed simplified that far.

I'm also aware there are some online tools and possibly some mods for this sort of thing. While those would also be interesting to know about, the preferred solution should be a calculation that someone with a non-specialized U.S. high school education (or less) could do in their head (or in one or two calculation steps on a basic calculator), given the above known variables, while playing the stock KSP game.

  • 1
    You can always stick your return configuration on an SRB, sling it up, de-couple, deploy chutes, and see what happens. You only need to see that the terminal velocity is low enough, you don't need to actually land. Then "revert to VAB" once you know - I think of these trial and revert exercises as "computer simulations". :-) Commented Nov 1, 2017 at 14:21

3 Answers 3


You're best bet is to err on the side of safety and use too many. The potential financial loss for chutes ($700 each for radial) is much less than the potential financial loss of your entire ship.

I'd love to be able to give you some simple formula like "Use 1 chute for ever x tons". But there is no simple formula. The complex formula can be calculated using the tool you've linked to.

That said, these are some excellent tips:

  1. Balance your lander - position your chutes so you land as flat as possible.
  2. Land on a flat surface - otherwise you could topple over.
  3. Keep your ships flat - pancakes don't topple over.
  4. Use modular girders Modular girder or i-beamsi beam instead of lander legslarge lander leg. Modular girders and i-beams have 6x higher impact tolerance (80m/s) than the strongest lander legs (12m/s). You can land a 1000 ton ship on Kerbin with a single chute if it can survive a 80m/s landing.
  5. Fire your rockets prior to landing to slow down your ship.

Good luck.

  • 3
    Regarding girders/beams: How does the impact get handled if they're attached to weaker components if only the girder touches down? For example, if I've got a girder that can take an 80 m/s impact attached to a fuel tank or other component that can only take 6 m/s, and I land on the girder at around 79 m/s, does the girder take all of the impact? Or will the weaker component be destroyed?
    – Iszi
    Commented Jul 29, 2014 at 15:57
  • 1
    @Iszi It looks like the impact stresses spread out from the initial impact component - girder takes 80, fuel tank takes 20. This can be mitigated by using multiple parts (girders strung together) or MOAR STRUTS!
    – Coomie
    Commented Jul 30, 2014 at 6:52
  • I use Nova Punch which has parts that can take 160 m/s impact. Using engines under the tanks, I've landed at 100+m/s without damage (that's near terminal velocity).
    – Coomie
    Commented Jul 30, 2014 at 6:55
  • Note that the landing legs have some springs/dampers built in, so they will not break immediately on an impact of greater than 12 m/s but instead try to cushion the shock. Only when they run out of suspension travel they will start to break. Commented Apr 6, 2023 at 16:15

As always and with everything in KSP - it's a long and bo0O0oring formula to calculate the amount of parachutes needed for successful landing. The formula, that I don't even bother to find out, but someone does, and they wrote a parachute calculator.

Please, use it here: http://ksp.freeiz.com/

EDITTED 14/07/31: As long as solution achieved by radial chutes, it is obvious, that the number of those should exceed the total stage tonnage by 33%. To easily calculate this without alt-tabbing and playing with your phone's calculator is to divide tonnage by 3 and add to tonnage. But if we are talking about "as safe as possible" - than just divide by 2 and add to tonnage. EXAMPLE: the parts' mass is 10t, 10/2 = 5 ... +10 = 15 (chutes)

Total Mass of components to be safely returned to Kerbin and Mass and Drag of each parachute model may be found there. And for the sake of the gameplay I'd suggest not to base your Impact Tolerance of weakest component assumption on any current bugs, that would probably be fixed (in a very distant future, but still), therefore exploiting them.

For placement of the parachutes it's best to start a separate ship project, start it with some lightweight part and attach the stage or disposable parts, then depending on the center of mass - arrange parachutes.

Forgive me my mistakes. Hope it helps.


My rules of thumb for parachute landing on Kerbin:

Total # Mk16 or Mk2r class Parachutes = 1/2 Landing (empty) Mass; 1 parachute per 2 tonnes.

Total # XL class Parachutes = 1/4 Landing Mass; 1 parachute per 4 tonnes.

The drogue parachute counts as an XL class for 4 tonnes but you should not use more than 2 of them for any size ship and never for more than 20% of tonnage of ship.

You do not need drogue parachute on ships less than 50 tonnes.

These rules give a touchdown speed of 6-7ms near sea level.

  • 1
    This doesn't work in version 1.4. Now it's more complicated, Mk16 and Mk2R are not equivalent, and spread angle on the radial chutes matter (more spread angle slows you down more). The calculator linked above is offline now. Commented Nov 2, 2018 at 9:44

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