Time to tandem terminal (from incidents)

[Andrewwhyte 1]

Re: [Lucky...] Tandem student and instructor injured, Cottonwood, Arizona, 7th May 2006 [In reply to] Quote | Reply


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By the time you figure you can't find it, you're probably near terminal amyway (esp if you have a heavy student),


I don't believe that is correct. Acceleration is constant regardless of weight, and if anything it should take slightly longer to reach terminal with a higher density as the terminal will be a higher.

P.S. I am commenting on the physics not how quickly it takes to find a drogue as I haven't got a clue on that front.

[Andrewwhyte 1]

Re: [nigel99] Tandem student and instructor injured, Cottonwood, Arizona, 7th May 2006 [In reply to] Quote | Reply


You're exactly right. It takes nearly 11 seconds to get to tandem terminal more than enough time to react to having no drogue.

[Andrewwhyte 1]

So I am confused. It takes about 12 seconds for a solo belly flyer to reach terminal; if it takes only 11 seconds for a tandem pair to reach tandem terminal (~50%) faster does that not imply a discontinuous second deriviative of speed/time?

[nigel99 616]

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So I am confused. It takes about 12 seconds for a solo belly flyer to reach terminal; if it takes only 11 seconds for a tandem pair to reach tandem terminal (~50%) faster does that not imply a discontinuous second deriviative of speed/time?

I am pretty interested in hearing a decent explanation of this. I guess that the drag coefficient may mean that you reach terminal at approximately the same time - implying that as you approach terminal your acceleration bleeds off (which is logical I suppose as you don't suddenly stop accelerating).

Experienced jumper - someone who has made mistakes more often than I have and lived.

[JohnRich 4]

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as you approach terminal your acceleration bleeds off (which is logical I suppose as you don't suddenly stop accelerating).

Correct. Most of the acceleration is in the first 5 seconds or so.

Look at the freefall table in an old log book, and add some new columns to that:

Seconds........: the number of seconds of freefall.

Feet per second: the number of feet fallen during that second.

Cumulative feet: the total number of feet fallen since exit.

Change FPS.....: feet-per-second change in fall rate over last second.

MPH............: fall rate at that second in miles per hour.


Feet per   Cumulative   Change

Seconds    second       feet       FPS      MPH

-------   --------   ----------   ------    ---

1         16          16        +16       11

2         46          62        +30       31

3         76         138        +30       52

4        104         242        +28       71

5        124         366        +20       85

6        138         504        +14       94

7        148         652        +10      101

8        156         808         +8      106

9        163         971         +7      111

10        167        1138         +4      114

11        171        1309         +4      117

12        174        1483         +3      120

13        174        1657          0      120

As you can see in the "Change FPS" column, the acceleration declines rapidly after 5 seconds.

What that chart would look like for a drogueless tandem, I don't know...

[strop45 0]

Acceleration is a constant only if you ignore drag/friction. As your speed increases this creates drag which effectively counters the gravitation force causing you to accelerate. When the drag is equal to the gravitation force, you no longer accelerate and are at terminal.

A tandem pair (without a drogue) take longer to get to terminal as they have greater mass and similar surface area.

The difference between stupidity and genius is that genius has its limits." -- Albert Einstein

[nigel99 616]

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Acceleration is a constant only if you ignore drag/friction. As your speed increases this creates drag which effectively counters the gravitation force causing you to accelerate. When the drag is equal to the gravitation force, you no longer accelerate and are at terminal.

A tandem pair (without a drogue) take longer to get to terminal as they have greater mass and similar surface area.

So on a practical note what timespan is there normally where a single free faller and a tandem pair are falling at the same rate - assuming no drogue? I guess as most tandems have their drogue out this is not likely to be something that alot of people have experienced?

Experienced jumper - someone who has made mistakes more often than I have and lived.

[Andrewwhyte 1]

I have heard people telling students that a tandem pair will get to terminal in x seconds without the drogue where x is a smaller number than 12. I wonder if they are talking about "regular" terminal (174ft/sec). Usually they are people not worth arguing with so I don't.
That said I JP just said it in the other thread and I do consider him worth arguing with.

[strop45 0]

Drag is a function of speed squared. So if a tandem is twice the weight of a single skydiver but with the same drag function, then you would expect the terminal velocity of the tandem to be the square root of 2 * terminal velocity of the single skydiver, i.e. 41% higher.

Similarly I expect that it would take roughly 40% longer to get to terminal.

The difference between stupidity and genius is that genius has its limits." -- Albert Einstein

[jurgencamps 0]

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So I am confused. It takes about 12 seconds for a solo belly flyer to reach terminal; if it takes only 11 seconds for a tandem pair to reach tandem terminal (~50%) faster does that not imply a discontinuous second deriviative of speed/time?

I was told during my tandem course by the tandemexaminator : 17 sec to terminal velocity for a drogueless freefall

[billeisele 130]

if the plane is going 90 horizontal and terminal, for example, is 120 vertical

how long does it take to reach terminal?

Give one city to the thugs so they can all live together. I vote for Chicago where they have strict gun laws.

[JohnRich 4]

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if the plane is going 90 horizontal and terminal, for example, is 120 vertical
how long does it take to reach terminal?

The plane's horizontal speed is irrelevant. Acceleration downward is the same if you jump out of a jet airplane or a hot air balloon.

[JohnMitchell 16]

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The plane's horizontal speed is irrelevant. Acceleration downward is the same if you jump out of a jet airplane or a hot air balloon.

Not really. There is the effect of drag coupling that makes the forward moving object accelerate downwards slower. Shoot a ping pong ball out of a sling shot and you'll see what I'm talking about.

That old "fact" about a fired bullet and a dropped bullet hitting the ground at the same time is false for the same reason.

[strop45 0]

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That old "fact" about a fired bullet and a dropped bullet hitting the ground at the same time is false for the same reason.

Can you provide any references or equation which describe how this works.

AFAIK, the calculations for downward speed of a bullet used in ballistics calculations don't allow for any such effect.

The difference between stupidity and genius is that genius has its limits." -- Albert Einstein

[JohnMitchell 16]

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Can you provide any references or equation which describe how this works.

AFAIK, the calculations for downward speed of a bullet used in ballistics calculations don't allow for any such effect.

No, but Prof. Kallend could.

I've never done calculations for ballistic tables. It's a very complicated process. I believe that was the army's first application for computers in the 1940's. Do you know if they simply use D=1/2 AT(squared) to calculate bullet drop? I really don't know the math, just some of the principles involved.

[pchapman 279]

I agree that the vertical acceleration is not entirely independent of what is going on horizontally.

Acceleration downward is the same in terms of what gravity is trying to do, but different in the end due to drag. The drag (and its horizontal and vertical components) will depend on the body angle and the direction of the relative wind at any given moment.

Properly, one has to calculate everything in two dimensional space as the jumper transitions through "the hill". It's not quite the same as assuming the jumper is belly to earth and falling straight down.

But at a first approximation the vertical-only method is probably OK. How much of a difference there is, that's an open question.

[JohnRich 4]

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The plane's horizontal speed is irrelevant. Acceleration downward is the same if you jump out of a jet airplane or a hot air balloon.

Not really. There is the effect of drag coupling that makes the forward moving object accelerate downwards slower. Shoot a ping pong ball out of a sling shot and you'll see what I'm talking about.

That old "fact" about a fired bullet and a dropped bullet hitting the ground at the same time is false for the same reason.

Please explain "drag coupling". Unless lift is generated by the horizontal speed, the fall rate is going to be the same.

For example, a skydiver tracking up the horizontal wind on exit, may indeed fall a bit slower than another exiting in a cannonball tuck. But I believe we should talk about all other things being equal here - we're talking about a given shape, with both horizontal speed, and no horizontal speed.

The ping pong ball may appear to be falling slower, but I think that's just an optical illusion from the long horizontal distance.

As for bullets, the same thing applies. If the gun barrel is parallel to the ground, a fired bullet will hit the ground at the same time as a dropped bullet. Just because the fired bullet travels much further in the same time span, doesn't mean that gravity is any different.

Now, gun barrels are actually pointed upward, relative to the straight-line sights, in order to allow shooting at longer distances. So you can't use that alignment in the experiment, because that lofts the bullet upward before it starts to descend. That's why I specify having the barrel parallel to the ground. And that would not be a normal mode of shooting, because since the sights are above the barrel, you would never hit what you were aiming at. In order for the bullet to intersect the line of sight, the barrel has to be pointed upward.

[billvon 3,110]

>There is the effect of drag coupling that makes the forward moving
>object accelerate downwards slower.

Initially there is no difference, and that effect is zero if the initial trajectory is perfectly horizontal. (Hence the "monkey falling out of the tree" scenario still works.) As the trajectory of the object begins to trend downwards, drag begins to take on a horizontal component as well as a vertical component, and thus the horizontal component of acceleration is less than it otherwise would be. (Note that the object itself is decelerating due to drag in any case.)

[strop45 0]

Looking at the prof's freefall calculation programme, there is a one second difference in freefall time for a change in plane speed of 200 kts - this may be 'drag coupling' or may just be a cumulative calculation difference.

Its hard for me to understand how any drag in a horizontal direction affects the drag in a vertical direction, except that the air is already moving near the interface between the skydiver and the air.

So IMHO its possible drag coupling does exist, however its effect for skydivers is so small it can be ignored.

The difference between stupidity and genius is that genius has its limits." -- Albert Einstein

[pchapman 279]

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Its hard for me to understand how any drag in a horizontal direction affects the drag in a vertical direction, except that the air is already moving near the interface between the skydiver and the air.

If one tried to calculate horizontal and vertical drag and acceleration independently, one would have to take body angle relative to the wind into account!

For a belly to earth jump from a balloon, one will be using the full belly to the wind area for the jumper to calculate the drag.

But if someone is jumping from a moving plane, and is for a moment at a 45 degree angle on the hill, one isn't going to be able to use the full belly to the wind area to calculate drag. One could try using a 45 degree angle to calculate the projected area.

The best thing to do would seem to be to do a full 2-D trajectory calculation, instead of trying to calculate the horizontal and vertical separately.

But would there be errors if the horizontal and vertical accelerations were calculated independently, if one already takes into account the changing body angle during the transition off the hill?

To do that "taking into account" one would have to calculate the area projected by the jumper in the horizontal and the vertical direction. And one would have to run both a horizontal and vertical calculation at the same time, as the area to be used at one microsecond of the calculation will depend on the trajectory (and thus body angle) both horizontally and vertically in the previous microsecond.

That's 'coupling' of a sort. But is that body angle issue the only 'coupling' there is, or is there still some other interaction that would cause errors?

I'm not seeing any clear reason for there to be any additional error. That's the part I'm not sure about thinking about this late at night. If there's no other error source, treating the axes independently works out accurately (as long as body angle is taken into account).

It can be asked whether a source of error would be that the drag coefficient may change somewhat depending on body angle to the wind. (Drag is a function of the area times the drag coefficient.) But the person is essentially belly to the wind all the time, so there wouldn't be any drag coefficient change. One would be using different areas for horizontal and vertical calculations, but keep the same coefficient.
(When the jumper is on a 45 degree trajectory, they aren't really at a 45 degree angle to the wind when calculating the vertical acceleration, even if one is using a projected vertical area of sin(45) = 70.7% of the full belly to earth area.)

[pchapman 279]

So I posed the issue whether there would still be errors if one calculated the horizontal and vertical speeds independently, as long as one took into account the body angle.

So there's already some "coupling" (as one poster termed it) because one has to do the horizontal calculations as well as the vertical ones, to figure out the body angle to plug into the vertical drag calculations. But for a given time slice being calculated (once one knows a given body angle), the vertical speed can still be calculated alone, without knowing the horizontal speed.

On further reflection I'm thinking that it is theoretically quite incorrect to calculate the vertical speed independently, due to the non linear relationship of drag and speed. So there is more "coupling" involved than I thought about last night.

This may or may not make a difference in real life big enough to care about -- one would have to run a simulation to see what the results really are. For the typical freefall tables, with typical aircraft speeds, it probably doesn't make enough of a difference for us to care whether one jumped out of a King Air or a C-182. But when trying to calculate the exact number of seconds to terminal velocity, errors would be introduced. (Of course one only approaches terminal velocity asymptotically, but one can come up with some way to determine what is close enough for our purposes.)


Here's the reason as I see it:

The calculation of acceleration from the forces on the jumper can accurately be calculated separately vertically and horizontally. Acceleration is force over mass, so acceleration scales linearly with the force. Therefore one can resolve the forces into 2 axes, do the calculations, and have correct results.

But this sort of thing isn't true when it comes to calculating the forces in the first place, that result from drag on the jumper. Drag forces vary quite closely with the velocity squared. That non linear term in the standard aeronautical drag equation makes it impossible to treat vertical acceleration separately.

Here's a simple look at numbers to make the point:
Say a jumper is on the hill, at 100 mph speed along his flight path, at an angle where he's moving forward 2 units for every 1 unit down. That works out to a 26.56 degree descent angle.

The horizontal speed component becomes about 89.4 mph, and the vertical about 44.7. Their contribution to the drag equation is based on those speeds squared, or 2000 vertical, 8000 horizontal.

Compare that to calculating things in a proper 2-dimensional manner. The actual speed on an angle is 100 mph, so the contribution to the drag equation is 10,000. When that is resolved into horizontal and vertical components, (where again one gets a vertical component 44.7% of the full value, and a horizontal one of 89. 4%), one has 4,470 vertical, and 8,940 horizontal.

Big difference.
The number in the drag equation is correctly 4,470 when resolving component of the drag in the vertical direction. But when we tried to treat horizontal and vertical separate, we only had 2,000. That's incorrect by over a factor of two in this particular example!

Therefore how fast a jumper accelerates downwards (with air to cause drag!) will depend on their forward speed. Not because the jumper might be tracking on the hill or some such thing, but because drag varies with the square of the velocity.

A jumper falling from a 727 will lose altitude slower than one falling from a balloon.

You could even have the situation where a jumper at a very high speed at a small descending angle may be slowing down vertically for a while - instead of accelerating faster towards the ground. Their vertical speed is nowhere near terminal, but with their very high air speed they are being slowed rapidly by high drag at high speed (hitting a wall of air).

So one cannot ignore the jumper's forward speed. Horizontal and forward speed are coupled.

Still, that doesn't answer the original time to tandem terminal question.

[strop45 0]

The thing I don't understand is how a horizontal component of drag results in a vertical component. Put another way, the cos 90 degree is zero. Just because there is lots of horizontal drag doesn't mean that there is any vertical drag.

In general you can treat velocity/acceleration as separate vectors and they have no effect on each other.

The difference between stupidity and genius is that genius has its limits." -- Albert Einstein

[kallend 2,146]

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if the plane is going 90 horizontal and terminal, for example, is 120 vertical
how long does it take to reach terminal?

The plane's horizontal speed is irrelevant. Acceleration downward is the same if you jump out of a jet airplane or a hot air balloon.

That is not true, since drag is a non-linear function of velocity. Hence there IS a coupling between horizontal component of velocity and vertical component of acceleration (albeit not a very large effect).

...

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

[kallend 2,146]

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The thing I don't understand is how a horizontal component of drag results in a vertical component. Put another way, the cos 90 degree is zero. Just because there is lots of horizontal drag doesn't mean that there is any vertical drag.

In general you can treat velocity/acceleration as separate vectors and they have no effect on each other.

But v^2 is not a vector, is it?

Only if drag is directly proportional to v (linear) is there no coupling. This would be the case at low Re where viscosity dominates.

...

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

[JohnRich 4]

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if the plane is going 90 horizontal and terminal, for example, is 120 vertical
how long does it take to reach terminal?

The plane's horizontal speed is irrelevant. Acceleration downward is the same if you jump out of a jet airplane or a hot air balloon.

That is not true, since drag is a non-linear function of velocity. Hence there IS a coupling between horizontal component of velocity and vertical component of acceleration (albeit not a very large effect).

Please explain. In layman's terms.