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woodpecker

Terminal velocity

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one poster tried to predict behavior based on High School physics. that frequently is Newtonian and neglects the effects of air.

when I fly or jump, I really like the effects of air. thus the H.S. physics model no longer applies.

I tried to build upon that model, to the extent required by most jumpers. (there is much I do not consult before jumping, string theory being one.)

I meant to inform at a level ALL could understand (while obviously risking offense via oversimplification). apologies if my approach annoyed.
DON'T PANIC
The lies in learning how to throw yourself at the ground and miss.
sloppy habits -> sloppy jumps -> injury or worse

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The formula is this:

Sum of Forces = Mass * Acceleration

Now the only forces acting on the body in freefall is the friction forces from the air flying over you body. The coeeficient of friction can vary based on your clothes, the density of the air, and your body positioning. You mass is you mass and the acceleration is 9.81 meters per second squared.

All that really does is get you in the ball park.

Also realize i have only had college pyisics required for my Engineering degree. I am not a BASE jumper and I don't have any jumpsother than some AFF skydives.


The reason the shot put hits the ground first is because it has a different coeeficient of friction compared to a tennis ball.
"... and I'll jump off that bridge when I get to it...."

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The reason the shot put hits the ground first is because it has a different coeeficient of friction compared to a tennis ball.



The shot put hits the ground first because it has a higher terminal velocity. So when the tennis ball reaches its terminal velocity and stops accelerating while the shot put continues to accelerate due to its higher mass. Even if the tennis ball was polished smooth to equal the coeficient of friction of the shot put, it'd still reach terminal velocity faster, simply because its terminal velocity is much slower than the heavy shot put.

It's time now! My time now! Give me mine. Give me my wings! - MJK

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You are right it is because it has a higher mass but then that acceleration is offset by the drag forces from friction. Which goes back to your coeficient of friction. I will draw a crude free body diagram tomorrow and show the Newton equation. We had to do a bunch of crap like that in the Calculus based Physics cousrse I took. I barely got an A but I had an easier Pyhsics prior to that one which helped me out in a bunch in getting the A.

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Where is yuri base when you need him.
I'm sure he will explain it and proof it with help of mathematical explanation of the phisical laws...



Hear, hear. Bringing the finest mental masturbation to the object near you. ;)

I wondered, too, if initial horizontal speed slows down your descent a bit. It turns out it does, but not too much.

Long story short - on a BASE jump, you reach terminal velocity about a second and a half sooner than on a jump from airplane. This is because the high initial drag on a plane jump has a vertical component which slows you down a little more.

If your terminal velocity is 120mph, you'll reach 95% of it (114mph) at 10.0s on a BASE jump and at 11.4s on a jump from airplane doing 100mph. This assumes that you maintain the same boxman body position - no lift, only drag - throughout. 11.4s is time to reach 114mph vertically; however, the total speed will reach 114mph at 10.6s (due to residual horizontal speed). So the difference is not really noticeable.

Skipping the lengthy derivations, for BASE jump the speed V follows this differential equation:

dV/dt = g*(1 - (V/Vt)^2)

where t is time, g = 9.8m/s^2 = 32ft/s^2 is acceleration of gravity, Vt = 120mph is terminal velocity. This equation can be solved analytically:

V = Vt*(e^(2*g*t/Vt) - 1)/(e^(2*g*t/Vt) + 1)

where e = 2.718281828459045...

(BTW, this formula can be used to derive freefall chart with a step less than 1s. Johhny? ;))

For an airplane jump, we have two equations for horizontal Vx and vertical Vy components of speed:

dVx/dt = - g*(V/Vt)^2*Vx/V
dVy/dt = g*(1 - (V/Vt)^2*Vy/V)

which can be solved numerically using finite differences, see the attached spreadsheet.

The only useful result of this masturbation is that if you want the softest opening possible on a hop-n-pop, wait 2 seconds before pitching (see how the green line, the total speed, makes a minimum of 85mph at 2s). This is based on 100mph jumprun (Twin Otter). For King Air, wait more. For Cessna, wait less.

Yuri
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acceleration due to gravity is a constant, period.



Very false. Gravity at sea level is different than gravity at the top of Mt. Everest.

***



a person at altitude of H meters above sea level experiences and acceleration due to gravity of
a= g*R2/(R+H)2

I dont do math.
but last time i split hairs that much, that equated to an insanely small mumber. as in a hundreth or thousandth of a (M)2 acceleration.

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Sum of Forces = Mass * Acceleration


true

to expand on what you wrote:
Mass * A(gravity) - aerodynamic forces = Mass * A(total)

understanding A(total) depends on how someone calculates the aerodynamic forces... it is not simple.

Yuri might be right, I do not know his starting assumptions.

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Now the only forces acting on the body in freefall is the friction forces from the air flying over you body. The coeeficient of friction can vary based on your clothes, the density of the air, and your body positioning. You mass is you mass and the acceleration is 9.81 meters per second squared.


overall, incorrect.

granted, friction plays a significant role in boundary layer conditions.

but shape typically impacts the resulting forces much more. that is why Cessnas use tapered fairings. they are much more costly to manufacture than a simple piece of tubing. but it makes a huge difference.

remember, the coefficient of friciton would be the same for a painted tube or a painted fairing. but the fairing has more surface area. defining forces only on friction would mean the faired strut would generate MORE drag than a simple tube.

as the fairing is for more costly to manufacture, and would represent greater drag, why do it?

the answer is the profile drag on the fairing is far less than that of a tube. shape matters.

edited to add
to clarify, the discussion above meant to explain why Cessna streamlined the wing support strut. it also explains why manufacturer moved away from wire cross-bracing. (just in case I mislead with my choice of terms.)
DON'T PANIC
The lies in learning how to throw yourself at the ground and miss.
sloppy habits -> sloppy jumps -> injury or worse

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Mass * A(gravity) - aerodynamic forces = Mass * A(total)



Correct. ma = mg - Drag. At terminal velocity (a = 0), Drag(Vt) = mg. Since the drag is proportional to the square of speed, at speed V drag will be Drag(V) = Drag(Vt)*(V/Vt)^2 = mg*(V/Vt)^2. So,

ma = mg - mg*(V/Vt)^2

a = dV/dt = g*(1 - (V/Vt)^2)

QED :)
This is a precise equation and it only assumes that you maintain the same boxman body position. The exponential solution above is a precise solution of this equation and can be used to generate precise freefall chart with arbitrary timestep and duration (the spreadsheet contains the freefall chart with 0.1s step to 20s).

For a jump with non-zero initial speed, there are two equations for the components of speed. They are derived similarly.
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This is a precise equation and it only assumes that you maintain the same boxman body position.


it looks like you created ratios of Bernoulli's dynamic pressure...
[sarcasm]
what about changes in density as you descend? what about compressibility effects?
[/sarcasm]

it's a nice little derivation to show the relationship between acceleration and velocity. did you do something as simple to come up with the speed calculations?
DON'T PANIC
The lies in learning how to throw yourself at the ground and miss.
sloppy habits -> sloppy jumps -> injury or worse

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did you do something as simple to come up with the speed calculations?



Acceleration is the derivative of speed: a = dV/dt. This gives us the differential equation for speed:

dV/dt = g*(1 - (V/Vt)^2)

Integrating

dV/(1 - (V/Vt)^2) = g*dt

gives us inverse hyperbolic tangent on the left side and g*t on the right:

Vt*arctanh(V/Vt) = g*t

or

(1/2)*ln((1 + V/Vt)/(1 - V/Vt)) = g*t/Vt

and finally

V(t) = Vt*(e^(2*g*t/Vt) - 1)/(e^(2*g*t/Vt) + 1)

Fun, heh? ;)
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All based upon the very simple hypothesis that drag is proportional to the square of the velocity and the coefficient of drag is invariant. Given the simple assumptions, yes, the solution is a closed analytic form. The end result in reality though is still 9 to 12 seconds depending on humidity, temperature, jumpsuit, jumper, and aerials. The numbers by Yuri_Base seem reasonable but I would question having any digits beyond the decimal point. Too much splittin' hairs and an unrealistic level of accuracy given the nature of the experimental system. I say let's just make a jump instead!
Looks like a death sandwich without the bread - Steve Deadman Morrell, BASE 174

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I say let's just make a jump instead!



ain't that what Yuri just said?

his variables include time and terminal velocity.
so all that stuff about shape, friction, humidity, etc. get incorporated into the terminal velocity.

guess we need someone all instrumented up to figure out that number. GO FOR IT!
DON'T PANIC
The lies in learning how to throw yourself at the ground and miss.
sloppy habits -> sloppy jumps -> injury or worse

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