Requesting some mathematical help from the technically minded...

[skysurfcam 0]

G'Day,

Since it's over 20 years since my last attendance at a maths class, I wonder if someone could help me out with a formula to calculate the force exerted on the base of a skysurf board in pounds per square inch (or kilos per square centimeter) at given speeds.

Specifically, I'm looking to see what a change speed from 110 - 120 -130 mph will do in term of increased pressure exerted on the board.

As always, thanks in advance,

Craig

P.S. Don't forget to show your workings...

Brother Wayward's rule of the day...
"Never ever ever go skydiving without going parachuting immediately afterwards."
100% PURE ADRENALENS

[billvon 3,076]

Easy. Figure out the exit weight. Divide by the area of the board in square inches. That's roughly the pressure at terminal.

[tr027 0]

Im thinking you'd have to know the aerodynamic drag coeeficient of the board, and ignore and drag due to the rider, but I also do not have an Aeronautical degree.

"The evil of the world is made possible by nothing but the sanction you give it. " -John Galt from Atlas Shrugged, 1957

[Squeak 17]

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Easy. Figure out the exit weight. Divide by the area of the board in square inches. That's roughly the pressure at terminal.

that don't sound correct, it would be the pressure if standing on the board stationary and suspended somehow. But at TV you have 120mph winds hitting the base of the board aswell.
doesn't the relative wind have an influence on the presre placed on the board??

You are not now, nor will you ever be, good enough to not die in this sport (Sparky)
My Life ROCKS!
How's yours doing?

[bob.dino 1]

If you're falling straight down the hole, it's vaguely right. There'll be all sorts of other funky effects related to the density of the air, vortices around the edge of the board, etc etc etc.

However, I think Bill's suggestion is a decent first approximation, and definitely better than a Wild Ass Guess.

Then again, on occasion I have been known to be wrong .

[billvon 3,076]

>But at TV you have 120mph winds hitting the base of the board aswell.

Right. And they are exerting EXACTLY your exit weight on the board (and your body, but that's minor if it's a large board) at terminal.

>doesn't the relative wind have an influence on the presre placed on the board??

Yes. It makes the pressure go up until terminal is reached.

[n23x 0]

Allright, I have to be at work in 4 hours, I can't sleep, so let's nerd it up with an example:

Let's examine a ball in freefall.

(in the case of this problem, we are given densities of the ball and the surrounding air, and a Reynolds number which we use to later calculate the Coefficient of drag via iteration)

I don't think I'll be drawing a freebody diagram, and we are just going to examine the two upwards forces, one due to the "bouyant force" of the ball displacing the air, and one due to drag incurred by air flowing past the ball.

Fb + Fd = X

Fb = Density of air * volume of air displaced by sphere * G

Fd = 0.5 * density of air * velocity^2 * Coefficient of Drag * Exposed Area

So for a given velocity and Cd, we should be able to determine the force applied by the Fd and Fb on the ball.

Pressure = Force / Area, you've solved for force X, you have the given area, solve for pressure.

In theory, this would allow you to get a guesstimate of the pressue on the exposed surface for a given velocity. However, since you don't really have numeric values, I really like Billvon's method for solving at a given terminal velocity.

It's late and this is probably wrong, so I leave it up here for everybody to get a good laugh. Besides, it's been 2 years since I've worked out a fluids problem, so screw you guys anyways.

.jim

"Don't touch my fucking Easter eggs, I'll be back monday." ~JTFC

[kallend 2,106]

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Easy. Figure out the exit weight. Divide by the area of the board in square inches. That's roughly the pressure at terminal.

He didn't say his speed was terminal though. He asked the effect of changing windspeed on the pressure. Maybe he asked the wrong question.

...

The only sure way to survive a canopy collision is not to have one.

[JayCam 0]

Yeah you would really have to firgure out how much upwards force the wind exerts on the board. It till be less than your weight otherwise you wouldnt be falling? make sense. Newtons third law I think, or second. Once of them lol. I know I am right up to here from my a level physics but I dont know how you would calculate how much force this was.

Someone with a qualification in fluid mechanics please take over lol.

J

[jakee 1,564]

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It till be less than your weight otherwise you wouldnt be falling?

If it was less than your weight you would be accelerating. Once you reach terminal forces up and down will be balanced, and you will no longer be accelerating.

Do you want to have an ideagasm?

[larsrulz 0]

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It till be less than your weight otherwise you wouldnt be falling?

If it was less than your weight you would be accelerating. Once you reach terminal forces up and down will be balanced, and you will no longer be accelerating.

Ding...simple physics.....force equals mass times acceleration. Forces in equilibrium mean there is no acceleration (i.e. terminal velocity); there is no for all those other details unless you want to determine explicitly what your terminal velocity will be.

I got a strong urge to fly, but I got no where to fly to. -PF

[billvon 3,076]

>He didn't say his speed was terminal though.

True. If he was still accelerating, the force on the board would be equal to the amount of 'normal' force he felt holding him up, which would equal his weight at terminal.

[pilotdave 0]

As you can see, clearly nobody has any clue.

The pressure won't likely be distributed evenly along the bottom of the board. So I guess the question is why do you want to know? Maybe theres a better question to ask to get the information you are looking for...

Dave

[JayCam 0]

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It till be less than your weight otherwise you wouldnt be falling?

If it was less than your weight you would be accelerating. Once you reach terminal forces up and down will be balanced, and you will no longer be accelerating.

Yes you are perfectly right. I apologise.

J

[Darshiva8 0]

Ok, I'm an aerospace nerd, so I couldn't resist, though fluids is not really my forte. Anyway, here goes. The basic equation you start with is Newton's sum of forces: sum(F)=weight+drag. Since weight is acting down, which we will set to positive and drag up, nad setting drag as equal to bv, where b is a coefficient and v is a velocity (this sums up drag's dependence on velocity). Weight then is equal to mass times the acceleration due to gravity, and the sum is equal to mass times acceleration at the given moment.

So now ma=mg-bv. As the body increases velocity as it falls, acceleration decreases, and the drag force approaches your weight.

If you are changing your body position so less surface area is exposed to the wind and therefore finding an equilibrium or "terminal velocity" for that position, then it is very easy to calculate the pressure by taking your weight and dividing it by whatever area is exposed to the airflow, so your board and any exposed appendages and such.

Either way it would probably be a safe assumption that you are fairly close to terminal velocity, so the key is to try to figure out how much area other than your board the wind is catching.

Granted the above won't give you an exact value, but you can at least get the gist of what pressures your board is seeing.

**************
For once you have tasted flight you will walk the earth with your eyes turned skywards, for there you have been and there you will long to return.
~Leonardo da Vinci~

[Sen.Blutarsky 0]

Hey Darshiva8, I realize that fluids are not your main gig, however, Skysurfcam got me thinking how to solve for pressure force exerted by a board as it is moving in the _horizontal direction_ relative to earth. There is a user-friendly viscosity relationship I was considering where the pressure force equals the area of a board times the ratio of velocity over the vertical separation (AGL) at a given time with this product being multiplied by a simple coefficient of viscosity. So, F=A*v/l*eta where eta is the viscosity of air at 20 degrees C, 0.018*10^-3 Pascal seconds. This should give pressure force in the direction perpendicular to vertical provided the board keeps moving in a constant horizontal direction, no? Not to hijack, just curious, thx.

Blutarsky 2008. No Prisoners!

[kallend 2,106]

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Ok, I'm an aerospace nerd, so I couldn't resist, though fluids is not really my forte. Anyway, here goes. The basic equation you start with is Newton's sum of forces: sum(F)=weight+drag. Since weight is acting down, which we will set to positive and drag up, nad setting drag as equal to bv, where b is a coefficient and v is a velocity (this sums up drag's dependence on velocity). Weight then is equal to mass times the acceleration due to gravity, and the sum is equal to mass times acceleration at the given moment.

So now ma=mg-bv. As the body increases velocity as it falls, acceleration decreases, and the drag force approaches your weight.

If you are changing your body position so less surface area is exposed to the wind and therefore finding an equilibrium or "terminal velocity" for that position, then it is very easy to calculate the pressure by taking your weight and dividing it by whatever area is exposed to the airflow, so your board and any exposed appendages and such.

Either way it would probably be a safe assumption that you are fairly close to terminal velocity, so the key is to try to figure out how much area other than your board the wind is catching.

Granted the above won't give you an exact value, but you can at least get the gist of what pressures your board is seeing.

At the Reynolds numbers in question, the force due to dynamic air pressure goes as v^2, not as v.

So, if I understand his question correctly, which was "Specifically, I'm looking to see what a change speed from 110 - 120 -130 mph will do in terms of increased pressure exerted on the board."

So he needs to know his terminal velocity Vt (from his protrack or similar), his weight W, and the projected area of his board A.

Then at terminal, pressure Pt = W/A approx. (only approx because it ignores drag from his body, etc).

then at at any other velocity v, the pressure will be

P(v) = ((Pt)*v^2)/(Vt)^2

...

The only sure way to survive a canopy collision is not to have one.

[Newtons2nd 0]

Billvon seems to have the only simple yet correct approach.

I am not guessing at this or late for work, this is how it works.

P=F/A with the force being your exit weight and the area being the board plus the body parts exposed (which does need to be accounted for considering the amount of error in the first place). It would be best to draw a quick sketch from a "whuffo's" point of view on the ground (with good binoculars). The exposed arms drawn in two dimensions is probably enough and although over shooting actual, will make up for the other small protrusions.

The most amount of pressure you could inflict upon the board is having it horizontal to the earth with max poss velocity. Any decrease in this would be due to more bodily protrusion and can simply be subtracted using the exposed surface area of whatever it is you are sticking out.

The most important thing to remember is that in most cases you will increase velocity by tilting the board in which case the pressure is now in vector form and far less (surprisingly) at even higher speeds.

So again simply subtract the area exposed as you decrease velocity. You should not use any square of anything unless you have a mode of power behind it to drive a board in the same plane to a higher velocity, which is obviously not what you are dealing with.

If you want to take it a step further--there is one more greater force, and that is when you decelerate from 130-110. This can simply be figured by using Newtons2nd (imagine that) and you would have to just use the change in velocity over the change in time and then add it to the pressure you already have figured.

Notice again that everything is adding and subtracting, there is no complicated anything with this math.

The other posts are correct if you have eliminated all variables and have more knowns, otherwise you will bring in more error than it attempts to eliminate.

[Sen.Blutarsky 0]

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...although over shooting actual, will make up for the other small protrusions.

Speak for yourself, kid


Government work is God's own work!

[Newtons2nd 0]

I am sorry to hear about your misfortune.

I guess I should have remembered that some of you older guys have fallen out of condition and protrude in all of the wrong areas.

You know they have programs for your kin.

Aerobics would be a good start.

I did want to add one more thing to my last post but can't seem to find an edit button:

I assume you know when calculating the pressure on a board while decelerating you use F=MA (force, mass, accel) and then use P=f/A (pressure, force, area)

p.s. look up PARSIMONY (science definition) it will make your mathematical life much easier

[kallend 2,106]

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Billvon seems to have the only simple yet correct approach.

I am not guessing at this or late for work, this is how it works.

P=F/A with the force being your exit weight and the area being the board plus the body parts exposed (which does need to be accounted for considering the amount of error in the first place). It would be best to draw a quick sketch from a "whuffo's" point of view on the ground (with good binoculars). The exposed arms drawn in two dimensions is probably enough and although over shooting actual, will make up for the other small protrusions.

The most amount of pressure you could inflict upon the board is having it horizontal to the earth with max poss velocity. Any decrease in this would be due to more bodily protrusion and can simply be subtracted using the exposed surface area of whatever it is you are sticking out.

The most important thing to remember is that in most cases you will increase velocity by tilting the board in which case the pressure is now in vector form and far less (surprisingly) at even higher speeds.

So again simply subtract the area exposed as you decrease velocity. You should not use any square of anything unless you have a mode of power behind it to drive a board in the same plane to a higher velocity, which is obviously not what you are dealing with.

If you want to take it a step further--there is one more greater force, and that is when you decelerate from 130-110. This can simply be figured by using Newtons2nd (imagine that) and you would have to just use the change in velocity over the change in time and then add it to the pressure you already have figured.

Notice again that everything is adding and subtracting, there is no complicated anything with this math.

The other posts are correct if you have eliminated all variables and have more knowns, otherwise you will bring in more error than it attempts to eliminate.

I don't think any of the "obviously"s in your response are at all obvious from the question - you assume too much.

The question was very simple, how does the pressure change as the speed is changed? You have not answered that at all except to assume an "obviously" which was not stated in the question.

At the Reynolds numbers in question, subsonic dynamic pressures change with v^2

...

The only sure way to survive a canopy collision is not to have one.

[tso-d_chris 0]

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I don't think any of the "obviously"s in your response are at all obvious from the question

Hmmm. I tend to have the same complaint about the authors of pretty much every college math book I've used.

("Clearly, this implies....")

[Newtons2nd 0]

What do you mean I haven't answered it?

The pressure changes when the protrusions assume some of the wind which will scrub speed.

You are looking way to far into this. When the speed changes think "why".

I don't care what you do, you cannot increase pressure on a board beyond terminal if it is horizontal and there is least possible drag from a body, unless you ac/decelerate.

If you mean that I didn't answer it "literally" then I now see why it is not obvious to you.

Besides the only "obvious" (one by the way) is after the statement that you do not have a mode of power behind the board causing additional force, that means a motor of some sort.

So since that is not obvious then I will say it more clearly:

To all skydivers:

You do not have a motor attached to your back when you are skydiving. I know that this should be obvious, but since it may not be, consider yourselves warned.

As far as the assumption, there was none, and I don't care to explain my background, as I was only trying to help not argue. ALthough I guess I ended up doing that just a bit too (obviously).

[tso-d_chris 0]

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P=F/A with the force being your exit weight and the area being the board plus the body parts exposed (which does need to be accounted for considering the amount of error in the first place). It would be best to draw a quick sketch from a "whuffo's" point of view on the ground (with good binoculars). The exposed arms drawn in two dimensions is probably enough and although over shooting actual, will make up for the other small protrusions.

First you say we shouldn't account for body parts exposed, and then you say we should express those body parts in two dimensions? Obviously, you can't have it both ways.

The only way for the skysurfer to accelerate after about 10 seconds is to change the terminal velocity. This is done by varying the exposed body surface area, or tilting the board, which will reduce the board's cross sectional area surface area, but also, more than likely, increase the surface area of the body that is exposed to the relative wind.

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So again simply subtract the area exposed as you decrease velocity. You should not use any square of anything unless you have a mode of power behind it to drive a board in the same plane to a higher velocity, which is obviously not what you are dealing with.

Here is where you went terribly wrong. You might be able to question the credibility of an Engineering Professor's explanation of Physics, but I have before me a genuine, up to date Physics book**, which uses skydivers as a specific example.

"For objects moving at high speeds through the air, such as airplanes, skydivers, cars and baseballs, the resistive force is approximately proportional to the square of the speed. In these situations, the magnitude of the resistive force can be expressed as

R=(1/2)*D*p*A*v^2
"[Serway, p. 164]
where R is the resistive force
D= the drag coefficient
p= density of the air
A=cross-sectional area measured in the plane perpendicular to its velocity
and v = the velocity

Obviously, you have to square the velocity, even without an engine on your back. Obviously, the Professor was right on with his critique of your explanation.

Work Cited:

Serway, Raymond A. and Jewitt, Jr; John W; PHYSICS for Scientists and Engineers; (C) 2004 Raymond A Serway. Published by Brooks/Cole Thomson Learning. ISBN number 0-534-40842-7

Used without permission under Fair Use guidelines.

[kallend 2,106]

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What do you mean I haven't answered it?

The pressure changes when the protrusions assume some of the wind which will scrub speed.

You are looking way to far into this. When the speed changes think "why".

I don't care what you do, you cannot increase pressure on a board beyond terminal if it is horizontal and there is least possible drag from a body, unless you ac/decelerate.

If you mean that I didn't answer it "literally" then I now see why it is not obvious to you.

Besides the only "obvious" (one by the way) is after the statement that you do not have a mode of power behind the board causing additional force, that means a motor of some sort.

So since that is not obvious then I will say it more clearly:

To all skydivers:

You do not have a motor attached to your back when you are skydiving. I know that this should be obvious, but since it may not be, consider yourselves warned.

As far as the assumption, there was none, and I don't care to explain my background, as I was only trying to help not argue. ALthough I guess I ended up doing that just a bit too (obviously).

The original question was "Specifically, I'm looking to see what a change speed from 110 - 120 -130 mph will do in term of increased pressure exerted on the board." There was nothing in the question pertaining as to how, why, or whether this speed change could be achieved.

And when I said you haven't answered it, I meant that you (meaning Newtons2nd) have not (meaning a negative) answered it (meaning provided a complete and correct response to the inquiry). Is that clear enough now?

All you've done is ramble on about additional sources of power, etc., which have nothing to do with the question that relates very specifically to the effect of speed on the pressure exerted on a board.

...

The only sure way to survive a canopy collision is not to have one.