Math help

[pBASEtobe 0]

Can someone help!?!
I’ve been curious to find out what my velocity and position would be after freefalling a certain amount of time. I want to be a little realistic and include drag forces due to wind resistance. I’ve come up with the following excel spreadsheet but am getting data that just doesn’t seem correct. Does it really take that long to get to 100mph? I’ve made some assumptions such as mass, drag coefficient, air density, and frontal area. I guessed coefficient of drag and area by assuming terminal velocity is between 110 and 115mph (ended up with 111mph).

Equation # 6 was solved for v using the quadratic formula, resulting in Equation # 1.

Any ideas? Did I do something wrong? Are my assumptions way off? Are these values correct and we just think/feel differently while actually in freefall?

[tymkoder 0]

CHICKEN MAN
BASE 954

[leroydb 0]

http://www.thebasepoint.com/
go to tools and you can compair your answers to theirs.. hope this helps

added BR delay chart

Freefall Speed Data


Freefall Time (sec) Speed (mph) Distance Fallen (ft)
1 10.91 16
2 31.36 62
3 51.81 138
4 70.91 242
5 84.54 366
6 94.09 504
7 100.91 652
8 Terminal 808
9 Terminal 971
10 Terminal 1138
11 Terminal 1309
12 Terminal 1485
13 Terminal 1661
14 Terminal 1837

Leroy


..I knew I was an unwanted baby when I saw my bath toys were a toaster and a radio...

[shunkka 0]

check this link
maybe it helphttp://polishbase.offheading.com/tabele.html.html

-------------------------
"jump, have fun, pull"

[460 0]

The chart is not quite correct. This chart was promoted by Mark Hewitt via Vertigo many years ago and has been taken as correct ever since.

The distances are not bad but the speeds are off.
The speed is v=acceleration*time, where acceleration due to gravity is 32feet per second^2, which is equal to 21.8mph per second^2. This acceleration decreases around 4 seconds due to air resistance effects. To calculate speed at 1 second, it's approximately 21mph, at 2 seconds, its 40mph or so, at 3 seconds its 60mph or so, at 4 seconds, maybe 75mph or so.

Chris, Ph.D.

Looks like a death sandwich without the bread - Steve Deadman Morrell, BASE 174

[pBASEtobe 0]

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The distances are not bad but the speeds are off.

I've used those speeds to calculate distance. First I used the speed to calculate the acceleration with the acceleration due to drag incorporated then I used that actual acceleration to calculate distance. d=0.5at^2.
If the speed was off then the distances would be too.

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The speed is v=acceleration*time, where acceleration due to gravity is 32feet per second^2, which is equal to 21.8mph per second^2.

At t=0sec but then decreases from then on out in proportion to the square of the velocity.

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...speed at 1 second, it's approximately 21mph

Yup, got it.

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at 2 seconds, its 40mph or so

I got 38.1

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at 3 seconds its 60mph or so

This is where it starts to go off, I got 51.6

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at 4 seconds, maybe 75mph or so

I get 61.1. Hmm, any help with the formulas then if you think that's off?

[460 0]

do you mean you calculated the speed given reasonable estimates of distances?

and i assume you are using the standard k*v^2 drag formula?

Looks like a death sandwich without the bread - Steve Deadman Morrell, BASE 174

[pBASEtobe 0]

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do you mean you calculated the speed given reasonable estimates of distances?

No, first I used a formula to calculate speed (no distance variable in speed formula). Then I used that speed to calculate distance. That's why I say the distance should be off if the speeds are off.

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and i assume you are using the standard k*v^2 drag formula?

Yes, force due to drag = 0.5CdArowv^2
Cd = coefficient of drag, A = frontal area, row = air density
There is a tab on the excel sheet labeled formulas. It shows all the formulas I used.

[wwarped 0]

V = at is invalid.
it only works when acceleration is constant.
since your acceleration is changing as the drag increases, you can not use this equation.

I believe the correct method is to integrate your acceleration as a function of time.

remember, once you reach terminal, you are in a "steady state" condition.
i.e. the forces of gravity = the forces of the air.
therefore, since a = 0 and V = at, then V = 0.

now that would be WAY cool...

DON'T PANIC
The lies in learning how to throw yourself at the ground and miss.
sloppy habits -> sloppy jumps -> injury or worse

[K763 0]

Tom, can we move this thread to the number weenies board?

Thanks,
K

Love ya Russel!!!

[base283 0]

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Tom, can we move this thread to the number weenies board?

I think you should pin it at the top for a while. With the new fly clothing and techniques that been developed, FF times/distances have radically changed and jumpers would have more insight on the potentials of their setup relative to a certain site and their flying style.
This is an opportune time for the scientists/engineers
to get it together.
take care,
space

[460 0]

What we need are measurements with an accelerometer from launch to deployment. That would solve many of these unknowns. As far as I know, Adam F. from non-existant CR has used accelerometers quite a bit in his canopy development. There is indeed a lot we can do if have *some* actual data and not just speculations from some old chart and models of wind resistance.

Looks like a death sandwich without the bread - Steve Deadman Morrell, BASE 174

[TomAiello 26]

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...Adam F. from non-existant CR...

Yo! Consolidated Rigging, Inc, is still in business.

See Adam's post here, in relevant part:

  Quote

Posted on Friday, November 5, 2004 - 4:26 pm:

Hi All,
Adam Filippino here.
Contrary to what some may have interpreted, Consolidated Rigging is still very much alive but only selling canopies now and not rigs.

-- Tom Aiello

Tom@SnakeRiverBASE.com
SnakeRiverBASE.com