How to calculate tandem fall rate factoring in drogue?

[3mpire 0]

If you were to calculate the drag of the drogue as negative weight, about how much would it be?

Meaning if the tandem instructor weights 175 pounds and the passenger weighs 175 pounds but the drogue creates enough drag so that the "relative" weight of the pair is 200 pounds not 350 (for example):

175 + 175 - drogue = relativeWeight

I understand there are a lot of factors that go into it but just roughly speaking (ballpark)?

[Dokeman 0]

different rig manufactures use different size drogues. And some manufacturers offer different size drogue for the same rig.

[3mpire 0]

Right, it would depend on the size of the drogue, perhaps different drogues of the same size are designed a little differently that would create different fall rates.

but there has to be some kind of calculation that goes into determining that. If you want to have the average tandem fall at about 120 MPH and the average tandem weighs say 350 pounds, then you would need the drogue to slow the pair down to that speed.

I could say that I weigh 190 lbs out the door and fall at 120 mph on my belly and a if tandem pair at 350 lbs with a drogue match my rate, the drogue is like minus 160 pounds.

What I wanted is to say have a 140 pound TI with a 115 pound passenger and quickly calculate their fall rate.

So taking my guess above, 140 + 115 - 160 = 95lbs which would probably fall at less than 100MPH... I don't know if that's an accurate formula for a ballpark guess though

[DougH 270]

To get any meaningful input from your formula you would need to take into account the build of the instructor and student.

4'8 175 lb female tandem student falls like a boulder compared to a 6'6 235 lb male student.

What about drag from jumpsuit versus bikini.

Also, I never took much in the way of physics in college, but aren't we dealing with fluid dynamics here? Does a heavier (denser) pair , generate more speed through the fluid, and therefore more drag from the drogue? I am guessing (blindly) that there is some curved relationship between tandem pair density, and drogue drag. My hunch is that it isn't linear, each additional 10 pounds doesn't equate in the same factor of speed?

"The restraining order says you're only allowed to touch me in freefall"
=P

[3mpire 0]

I suck at math. what i've outlined is about as far as I'd take it ;)

I wouldn't be surprised if it isn't linear, and all the other variables (jumper build, suit, etc.) decrease accuracy. But you could probably eyeball it and give a guess with a +/- MPH range.

Just like you can argue from a physics perspective that ground speed doesn't "matter" in calculating exit separation, it's a useful easy to use number to help you come up with a rough calculation that is "good enough"

[crazyboy 0]

The german parachute association made tests with a drogue chute in a wind tunnel some years ago.

Drogue bridle lenght: 190", speed 120mph, average pull force: 180 lbs
IIRC they used a Vector drogue chute.

If your parachute fails to open, remember you have the rest of your live to fix it.

[3mpire 0]

That's cool that they did an experiment like that. That isn't too far off my spitball guess though as I thought about it I didn't include the weight of the tandem rig in my math.

I'm guessing a tandem rig weighs... 60 pounds?

[Bluhdow 31]

3mpire

Right, it would depend on the size of the drogue, perhaps different drogues of the same size are designed a little differently that would create different fall rates.

but there has to be some kind of calculation that goes into determining that. If you want to have the average tandem fall at about 120 MPH and the average tandem weighs say 350 pounds, then you would need the drogue to slow the pair down to that speed.

I could say that I weigh 190 lbs out the door and fall at 120 mph on my belly and a if tandem pair at 350 lbs with a drogue match my rate, the drogue is like minus 160 pounds.

What I wanted is to say have a 140 pound TI with a 115 pound passenger and quickly calculate their fall rate.

So taking my guess above, 140 + 115 - 160 = 95lbs which would probably fall at less than 100MPH... I don't know if that's an accurate formula for a ballpark guess though

Translation: "I want to jump with my girlfriend on her tandem. How much weight should I wear?"

Let me know what you find out. I'm also interested in getting laid.

Apex BASE
#1816

[3mpire 0]

What piqued my interest was talking with some TIs about fall rates and how floaty a light TI + light passenger is. The more I thought about it the more I wondered... well exactly HOW floaty? Can you put a number on it? etc.

Quote

Let me know what you find out. I'm also interested in getting laid.

Check that other thread for dating advice, I've got nothing for ya--I'm a family man.

If you want advice on how to child proof your studio apartment (you can't, move into a bigger place) or what car seat works best in a hatchback (there aren't any, buy a bigger car), I'm your man.

Edited to add: upon further reflection, here is some dating advice--if you need to ADD weight to keep up with your date you're doing it wrong if you need camera wings to slow down, you're on the right track

[Bluhdow 31]

3mpire

Edited to add: upon further reflection, here is some dating advice--if you need to ADD weight to keep up with your date you're doing it wrong if you need camera wings to slow down, you're on the right track

Winner!

Apex BASE
#1816

[excaza 1]

Assuming I did my math right, which is a tall order this late in the day, this should be good for a spitball of the terminal velocity.

m here is the total mass, which would include TI, student, and rig. g would be gravitational acceleration, which is around 32 ft/s^2. Unless you know the mass of everything in slugs, you can use the total weight in place of mg.

Cd and A are the respective drag coefficients and projected areas for the tandem pair (TP) and the drogue (D). I'd use a drag coefficient of 1.0-1.1 for the tandem pair and something like 0.75 - 0.85 for the drag coefficient of a flat circular parachute.

EDIT: I forgot the rho (air density) in the denominator of the final equation. Whoops!

[3mpire 0]

I see Greek. So I will assume that your equation is more accurate than mine

As the total mass of the tandem increases, would the fall rate change at a constant rate or would there be a bend in that graph?

[excaza 1]

3mpire

I see Greek. So I will assume that your equation is more accurate than mine

terminal velocity and instead of one body, we have two (TI/Passenger and drogue). When we're at terminal velocity, the force of gravity (our weight) is equal to the force of the wind pushing on us (drag). Skydivers are pretty unique in that we can achieve a very wide range of drag coefficients and projected areas, and therefore can get the wide range of speeds you see between someone starfishing and someone going head down.

3mpire

As the total mass of the tandem increases, would the fall rate change at a constant rate or would there be a bend in that graph?

Technically, yes there will be a bend. Assuming you're keeping the other variables constant (in real life they won't be, but we're just spitballing here!), it will look like the plot of y = sqrt(x) (or x = y^2). However, since we're far away from 0, our graph will look something like what I've attached (ignore the speed values, I just made up numbers for the constants).

So it's not technically linear, but it looks pretty linear (as much as saying that makes me feel dirty inside )

[costanza 0]

god damn you and your slugs.

[excaza 1]

I could have used slinches

[JohnSherman 1]

The drogue adds to the square footage available for drag. The Tandem pair has about "X" Sq. Ft. The drogue is responsible for some of it. Your question is how much.
In order to know what a drogue drags you must know its effective square footage. The Effective Sq. footage is the plan form sq. ft. times the Drag Coefficient.
In order to know the Drag coefficient of the device it must be tested. Drogues generally have a Cd of .20 or so. They are low drag so as to minimize oscillation.
We have done several instrumented test where we don't deploy the drogue until half way through the FF. We can analyze the data and conclude the effect of the drogue by equalizing the "Q" and observing the difference on Effective Sq. Ft. That difference is the effect of the drogue.
This results in about 90 pounds of drag with 2 guys weighing 170 pound each and a rig weight of 40 pounds.

[3mpire 0]

Thanks for sharing that, I was assuming that this type of testing had to be common with tandem system manufacturers. That's really interesting to me.

What affect does an oscillating drogue have on the tandem pair?

I know that for reserves oscillating pilot chutes are good because it helps to, as I understand it, catch clean air and wiggle the reserve out?

[erdnarob 1]

The easiest way is to follow a tandem pair and be equiped with a Protrack or similar device giving the fall rate. Aerodynamics is a complex science and as you said, many factors contribute to get a result. This is why even aerodynamicists validate their findings using a wind tunnel.

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

[excaza 1]

Whoa, my drogue Cd guess was way off, thanks John!

[yuri_base 1]

It's really easy to estimate the drag of the drogue if you measure sustained fallrate with and without the drogue. Ask jumpers doing a practice tandem jump to do this for you using any device that can measure fallrate (Viso, etc.)

If without the drogue sustained fallrate is V1, then W = Kd1*V1^2, where W is total weight, and Kd1 is the "magic" coefficient of drag from Wingsuit Equations.

If with the drogue sustained fallrate is V2, then W = Kd1*V2^2 + Ddrogue, Ddrogue is the drag due to the drogue.

So,

Kd1 = W/V1^2,

W = W*(V2/V1)^2 + Ddrogue

Finally,

Ddrogue = W*(1 - (V2/V1)^2)

For example, if W = 170+170 + 40 = 380lbs, V1 = 150mph, V2 = 120mph, then

Ddrogue = 380*(1 - (120/150)^2) = 137lbs

For purely theoretical estimate, without any measurements, we can estimate the drogue-less fallrate of the tandem pair as standard fallrate 120mph times square root of 2 (since drag is proportional to the square of speed, and weight is increased by a factor of 2). Then (V2/V1)^2 = 1/2, and Ddrogue = W/2, which for the total weight 380lbs gives Ddrogue = 190lbs. But with actually measured V1 and V2, the estimate will be much more precise.

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