Terminal Velocity on Eve... survivable?

[Quasar]

Had a rather... interesting... experience with my first lander on Eve.

I brought it in a bit too fast and had my chutes torn off: I'm sure a lot of people have had that problem. But where it got interesting was how I reacted. I had no thrusters on the lander so, in the absence of any other options, I EVA'd the pilot, dived off the lander, and started thrusting upwards with the EVA pack.

I survived. Bounced, but survived. Moments later my lander nearly crashed on top of me.

Admittedly being stranded on Eve in your EVA suit isn't a particularly noble ending to the mission, but what I'm wondering is... how did I survive? Was it just a bug? Or is the air-pressure on Eve so high that it reduces a Kerbals terminal velocity just enough for the EVA packs thrusters to make a fall from orbit survivable?

Anyone up to skydiving from Even orbit to find out?

[Markus Reese]

I think it is the increased air resistance, and is the gravity different? Less gravity will mean the better pack effectiveness.

[Ababwa]

Eve has more gravity I think. I had something similar to that happen once, I had this big ol' rocket making its way through the atmosphere when it decided to go crazy and start spinning. It got closer and closer to the ground and I was expecting it to just completely blow up upon impact but it ricochet and ended up thrusters-down and began making its way upwards again. A few moments later it crashed again and blew up for good.

[BurningSky93]

Less gravity will mean the better pack effectiveness.

Eve has 1.7x Kerbin's surface gravity (1.7 G).

Terminal velocity is determined by:

V(t) = ((2mg)/(rho A C))^0.5

Where V(t) is terminal velocity, m is mass of the object, g is acceleration due to gravity, rho is density of the fluid through which the object is falling (in this case the atmosphere), A is the objects projected area and C is the object's drag coefficient.

In this case, m, rho, A and C can be assumed to be equal on Eve and Kerbin (the Kerbinaut's mass, falling area and coefficient of drag won't change.

So, if Eve has an acceleration due to free fall of 1.7x Kerbin and an atmospheric density 5x greater than Kerbin, and terminal velocity is directly proportional to the root of acceleration due to gravity and inversely proportional to the root of the atmospheric density, then terminal velocity on Eve is

(1.7/5)^0.5 = 0.58 times that on Kerbin.

Hopefully my math is correct, do please correct me if I have made a mistake though

[dikkjo]

Atmosphere density on Eve is much bigger than Kerbin, so rho change indeed and so do the drag. On the wiki you may find the exact values, if you want to correct your maths

[Ronox]

Haha I've had the same happen to me! I had run out of fuel after my intercept burn, ended up being maybe 10k from eve's atmo so I eva'd and reentered with my kerman who to my surprise bounced and lived!

[JustinKingr]

I did practically the same thing on Kerbin today when my rocket didn't make it into orbit. Sadly he didn't quite make it and exploded the second time he hit the ground...

Edited September 26, 2012 by JustinKingr

[BurningSky93]

so rho change indeed and so do the drag

I did take that into account, I must've not been paying attention when I wrote "rho... can be assumed to be equal on Eve and Kerbin", but my maths does take a change in rho into account.

Also i think my math is a bit off because Eve's atmo isn't actually 5x the density of Kerbin's. It was revealed to be that during development but I dont think that is the actual value.

I think my actual calculation is sound, even if the actual variables aren't.