How do I explain the differences in freefall rates?

[Hellis 0]

bdenny20

Hi.

So I went up to my folks' place for Christmas. This is the first time my entire family has been together since I picked up skydiving. While I was talking to my brother about it, I mentioned that because I'm a larger and heavier guy, I've been having a little trouble flying with other people because...well, most of you fuckers are small and I have some difficulty slowing down.

Now, real quick. He comes from an scientific background. Chemistry, but that path included lots of courses from other areas of focus, including physics. He immediately goes off saying how that shouldn't be a problem, gravity effects everyone the same, fall rates should be equal, yadda yadda yadda you know the type.

I'm not from a scientific background, let's put it at that. I like flying airplanes and as of recent, jumping out of them. I wasn't able to convince him...so I pulled up a video of a recent jump with me an one other small chick. It shows us docked in a two way belly, and as soon as we separate, I take off like a rock.

He's perplexed. I know of the concepts of air resistance and how I'm still new and my form isn't spot on, but I'm not smart enough to come up with an educated response....How do I explain to non-skydivers that us fattys have a hard time keeping up with lighter peeps?

Maybe he was thinking of how things fall if there is no air?
If you remove the air resistance from it, he is correct.
And most likely he remembers this from school because it's one of those things the teachers 'outsmarts' the class with.

I believe Mythbusters tested it in one of thier episodes.

[erdnarob 1]

I noticed this error. By doubing the area of the face of a cube you multiply the edge of the cube by (SQRT 2) therefore the new volume becomes the old volume x (SQRT 2) ^3) = 2.828 x the old volume as you mention it.

But don't worry about Peter Maths. Just a distraction.

Learn from others mistakes, you will never live long enough to make them all.

[erdnarob 1]

Hi ryoder

Here is a nice explanation of what is going on here. Thanks

But since nobody is perfect (including myself) you have forgotten a parenthesis at the very end of your formula. You have 5 parentheses instead of 6. This to make sure the SQRT applies to the whole expression.

Note : are you sure that you have to multiply (m*a) by 2 ???
(m*a) = the force of gravity or weight...anyway, that doesn't change the logic following

Learn from others mistakes, you will never live long enough to make them all.

[ryoder 1,590]

erdnarob

Hi ryoder

Here is a nice explanation of what is going on here. Thanks

But since nobody is perfect (including myself) you have forgotten a parenthesis at the very end of your formula. You have 5 parentheses instead of 6. This to make sure the SQRT applies to the whole expression.

Note : are you sure that you have to multiply (m*a) by 2 ???
(m*a) = the force of gravity or weight...anyway, that doesn't change the logic following

Yes, due to a copy/paste mishap from composing in a terminal window, the closing parenthesis got lost, but I didn't notice until it was too late to edit it.

And yes the "2" belongs in there. Look around on the net and you will find the same equation. In some cases the "m * a" is replaced by a "W" for weight).

"There are only three things of value: younger women, faster airplanes, and bigger crocodiles" - Arthur Jones.

[Sincy78 1]

So there really is no such thing as terminal velocity. For a constant CdA, rho is changing with altitude, therefore so is velocity. It should be called terminal dynamic pressure.

[JohnMitchell 16]

Sincy78

So there really is no such thing as terminal velocity. For a constant CdA, rho is changing with altitude, therefore so is velocity. It should be called terminal dynamic pressure.

Very true, but think of it as a comparative point on the curve.
Or maybe wind tunnel speed?