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.
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:
- Balance your lander – position your chutes so you land as flat as possible.
- Land on a flat surface – otherwise you could topple over.
- Keep your ships flat – pancakes don’t topple over.
- Use modular girders or i-beams instead of lander legs. 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.
- Fire your rockets prior to landing to slow down your ship.
Source : Link , Question Author : Iszi , Answer Author : Coomie