P.C.'s Drag Force Table

[guest]

Seen that in a recent post (stieviboy May-21-00, 10:48 AM (PST) "opening speed") somebody (space) was asking for P.C.’s drag forces, I thought about posting this one.
The hereinafter reported formulas for freefall were reported on the BASE Board last year by a jumper (Basetroll ?), I implemented them on an Excel spreadsheet, plus I added the formula for air pressure and for resistance force caused by air pressure on a given surface, finally generating the hereinafter reported table which expresses the theoretical drag force exerted by a given ZP P.C. at a given airspeed. What I got are numbers that make sense. So here we go.
Please, forgive me for the form of present post, but I am not so clever in using the BASE Board editor for writing down formulas and tables.

ρair = 1.225 kg/m³ air density at sea level

cw = shape coefficient of skydiver body
mb = overall jumper body weight &l;kg&r;
A = jumper exposed surface &l;m²&r;

cw = 0.28 frog position
cw = 0.22 tracking position
cw = 0.18 head down position

g = 9.8 m/s² gravity acceleration
k = 0.5 · ρair · cw · A resistance coefficient &l;kg/m&r;

VL = sqrt(g · mb / k) terminal velocity &l;m/s&r;


Assumed a jumper of overall body weight mb = 80 kg in frog position (cw = 0,28 ) and of body surface A = 1.8 m², we shall have then:

k = 0.3087 kg/m
VL = 50.4 m/s

helping factor q = exp(VL · (2k/mb) · t)

a(t) = 4 · g· ( q / ((q 1)²) ) acceleration &l;m/s²&r;
v(t) = VL · ((q - 1)/(q 1)) velocity &l;m/s&r; or
v(t) = VL · tanh(VL · (k/m) · t) velocity &l;m/s&r;
s(t) = (VL²/(2 g)) · ln( ((q 1)²)/(4 q) ) distance &l;m&r;

P = ½ · ρair · v² air pressure &l;N/m²&r; (where v² is the square of velocity)

F(N) = P · APC where APC is the area of ZP P.C. in m²
F(kg) = F(N)/9.8
F(lb) = F(kg)/0.4536
m = 0.3048 · ft
ft = 3.2808 · m
m = 0.0254 · in

where m is meter, t is time expressed in seconds, s is second, kg is kilo, N is Newton, lb is pound, ft is feet, in is inch, ln is the natural logarithm, tanh is the hyperbolic tangent, sqrt is the square root, exp( x ) is what comes from e elevated to x (where e is the natural logarithm base (e = 2.718))


In the hereinafter reported table the drag force F is expressed in kg and is reported in the column headed by the diameter of ZP P.C. expressed in inches.


P. C.’s Drag Force Table
t s(t) v(t) a(t) P 48" 46" 45" 42" 40" 38" 36" 35" 32" 28"
&l;s&r; &l;m&r; &l;m/s&r; &l;m/s²&r; &l;N/m²&r; F F F F F F F F F F
0 0 0 9.8 0 0 0 0 0 0 0 0 0 0 0
0.5 1 5 9.7 15 1.7 1.6 1.5 1.3 1.2 1.1 1 0.9 0.8 0.6
1 5 10 9.4 57 6.8 6.3 6.0 5.2 4.7 4.3 3.8 3.6 3.0 2.3
1.5 11 14 9.0 125 15 14 13 11 10 9 8 8 7 5
2 19 19 8.5 213 25 23 22 19 18 16 14 14 11 9
2.5 29 23 7.8 317 38 35 33 29 26 24 21 20 17 13
3 42 26 7.1 429 51 47 45 39 35 32 29 27 23 17
3.5 56 30 6.4 545 65 60 57 50 45 41 37 35 29 22
4 72 33 5.6 660 79 72 69 60 55 49 44 42 35 27
4.5 89 35 4.9 771 92 84 80 70 64 58 52 49 41 31
5 107 38 4.3 874 104 96 92 80 72 65 59 55 46 35
6 147 41 3.2 1054 126 115 110 96 87 79 71 67 56 43
7 190 44 2.3 1196 142 131 125 109 99 89 80 76 63 48
8 235 46 1.6 1302 155 142 136 119 108 97 87 82 69 53
9 282 47 1.1 1379 164 151 144 126 114 103 92 87 73 56
10 330 48 0.8 1433 171 157 150 131 119 107 96 91 76 58
11 378 49 0.5 1472 175 161 154 134 122 110 99 93 78 60
12 428 49 0.4 1498 178 164 157 137 124 112 100 95 79 61
13 477 50 0.2 1516 181 166 159 138 125 113 102 96 80 61
14 527 50 0.2 1529 182 167 160 139 126 114 102 97 81 62
15 577 50 0.1 1537 183 168 161 140 127 115 103 97 81 62
16 627 50 0.1 1543 184 169 162 141 128 115 103 98 82 63
17 677 50 0.1 1547 184 169 162 141 128 116 104 98 82 63
18 728 50 0.04 1550 185 170 162 141 128 116 104 98 82 63
19 778 50 0.02 1552 185 170 162 142 128 116 104 98 82 63
20 828 50 0.02 1553 185 170 163 142 128 116 104 98 82 63



What I would like to point out about hereinabove reported table is that the drag force values are theoretical drag force values, in the sense that they are actuall drag force values if, and only if, you have an infinitely rigid disc (i.e., steel plate) put in an airstream (with its axis perfectly parallel to the airstream itself) at that given speed and having no burble in front of it.
So the most experienced BASE jumpers can (being perfectly right) argue that the reported drag force values must be diminished (of how much, I do not know !) taking into account that: 1) in front of the PC there is a small burble (jumper’s body), as smaller as longer the bridle is; 2) the fabric of PC is not ZP (Zero Porosity); 3) the actual diameter of inflated PC is different (=narrower) than the diameter of PC put at rest on the floor, depending on PC manufacture, etc.; 4) the axis of PC could be not parallel to air stream depending on jumper’s body position during freefall; 5) other factors.
So, please, equipment manufacturers or other experts can find and tell us the “diminishing factor” to be divided by above drag force values to generate the actual drag force values.

What above could be exact, not exact, close from being exact, far from being exact, I do not know.

But.

Stated all what above, one thing is for sure. The ratios between hereinabove reported drag force values are valid whichever the “diminishing factor” could be. Being the drag force proportional to P.C.’s surface, at any given air speed it is always true that a 48” has got 3 times the drag force than a 28”, and a 48” has got twice the drag force than a 35”.

(being the ‘at rest’ areas of above P.C.’s as follows: 48”: area = 1.17 m²; 35”: area = 0.62 m²; 28”: area = 0.40 m²).

All above said and done, I wish blue skies to you all

Andrea Checchia

[guest]

Andrea:

If I am reading this table correctly, a 28" PC at 1 second exerts more force (2.3) than a 48" PC at .5 seconds (1.7). Further the same 28" PC at 3 seconds exerts 10 times as much force (17) as the 48" PC at .5 seconds.

Am I reading it correctly?

If so, the results are startling, to say the least. The math appears to contradict the received wisdom of our sport.

However, some experiences do seem to bear the math out. For example, the earlier thread about openings with 42" and 36" PC's appearing to occur at the same speed would bear out the idea that the size of the PC matters a great deal less than we often think.

Another thought is that perhaps the "diminishing factor" grows larger over time, or is related to PC diameter in a non-linear fashion (effecting a 28" PC 10 times as much as a 48", or effecting a 3 second deployment 10 times as much as a .5 second deployment).

If the math is accurate, and the theory is even close to reality, then perhaps it really isn't the size of the PC, but the motion of the jumper that matters...

The bottom line is: am I understanding the math correctly?

--Tom Aiello

[base615]

Granted I'm not the sharpest tool in the box and I haven't done much in the way of maths since college but I read the table the same way as Tom did.

There's got to be a few others out there who are as lost as me when it comes to hyperbolic tangents and all that good stuff, so if anybody with a better grasp of maths/physics than me could put it into simpler terms you would be my hero!!!

Cheers,

Craig

[imported_mknutson]

I have not studied this enough yet but....
Isn't it true that the larger the pilot chute, the longer it actually takes to inflate? If this is true then the chart would be telling you that within the 1st second, a 46" PC is still inflating and not excerting as much force as a 28" at 1 second. The force curve for each PC is not linear and will be different from the other sizes.

--
Thanks

Mick Knutson
BLiNC Magazine

"Everything you ever wanted to know about BASE Jumping, but didn't know who to ask."
--

[squirrel]

finally, a discussion worthy of us Engineering/BASE jumpers :). the table is correct, if read left to right for each time, you'll see that the force decreases as the pilot chute size decreases. My college aerodynamics professor always said that calculations like these are only theoretical and need to be verified experimentally (he would be so proud). all you would need is a pick-up truck and a fish scale. if you read the force on the fish-scale in 5 mph increments, you could then derive a force vs. speed curve for each pilot chute. then, it's a straight forward math problem to correlate speed to the delay taken. remember, calculations don't take turbulence and mesh, and bridles into consideration. i have a 0-30 lb fish scale here at work, i'll try it this weekend and see how the #'s relate....

[guest]

GUys,
one thing is sure, i saw what i saw on video and a smaller pc is going faster open. it needs less air to inflate, it's logic. the only difference between a big one or a small one is only the "force" for opening the harnas, but with a small one u already have enough force to open it after 1 sec. so... make a jump like we do and u will see. Enjoy your 2-ways...

[guest]

Please do not insinuate that everyone should use a smaller PC because your video showed a given performance in your situation. I have published recommended PC sizes for given freefalls. All the manufacturers accept these standards. The real issue in these threads is theoretical performance, not recommended uses. The person that understands the advanced topics displayed here, could benefit from such discussion. A person without such knowledge could have problems and get confused.

Each situation has it's own characteristics.

[guest]

of course mike.
i think everybody have to know that because:
When we did this 2 way, we were(and we still) thinking like all the manufactures said(size/pc/sec), and we were sure that my mate will be open "under" me, and what's the surpise same level open and 1/2 meter horizontal distance from each other, so, i posted this to let people know that "could" happen "too". and this everybody had the right to know that, right.

I hope i didn't made a mistake with posting that, just let people know from mistakes...it will help.

c'ya

[guest]

Tom,

You're reading the table correctly and the table is correct.

Why are you surprised that at low air speed a large PC can have a lower drag force than a smaller PC at higher air speed. I think it make sense, it depends on the PC sizes/air speed. The opposite example is a 48"/6s force (126kg) is twice the force of a 28" even at twice the ff time (28"/12s=61kg).
In your example, 48" at 0.5s and 28" at 1s, one could think that you better have to use the second solution. First of all, in the first solution (throwing a 48" at 0.5s) the opening process starts earlier and higher. If you consider the extrem case that your 48" is not inflated between 0.5s and 1s, at 1s you are in the same situation than the second solution but with a 48" PC which you can see it is significantly a better solution.
The drag force of the PC is not constant during opening and is certainly not equal to drag force of a PC fully inflated in an airflow equal to the ff speed of the jumper at the time he throw the PC. To understand, we can consider that the pull forces of the pack job between pin/velcro pulling and start of canopy pressurization are about the same, regardless of the opening speed or altitude. During this stage the PC drag force will tend to be equal to the pack job force. So if the drag force is higher, the PC will slow down. If the drag force is lower (than the pack job pull), the PC will speed up. This means that with a larger PC, this will happens at lower air speed. Just imagine that an infinite size PC sticks to the air as soon as you throw it. This simply means that it will take less altitude. This is not so true because the slow down and speed up times are not negligible and it reduce this effect. In addition, the pack job pull (on the PC side) is less in a high air flow. This is one reason why the PC sizes in the earlier thread does not seem to be so important. Another reason is that during the next stage (pressurization), a higer airspeed contribute to the cells inflation.
All that to say that the drag force calculated in this table is not at all (even in theory) the drag force during openning. This does not mean that they are false but must be used for what they are.

What you have to keep in mind is that those values are theorical and as Andrea said, we are considering a perfectly pressured PC. The reality is very different especially at 0.5s or 1s, and you surely can forget the first line as valid data.

The misunderstanding usually cames from the fact that people often confuse speed, time and altitude. A fast opening does not mean a high opening. With a 48" PC and 0.5s delay, the opening time will be for sure slower than with a 28"/10s, but what about the open altitude?

Another point is the parameters value used for the calculation of the table. They can be discussed but if you use the result on a "relative" point of view, this table is usefull. However, be carefull because if you change one parameter (considering wingsuit for example), the ratio between values of the table can change significantly.

I do not agree with the value of cw and A parameters but I agree with the product so it does not changet the table data.

[space]

Mick is absolutely correct in "Each situation has it's own characteristics."
Each manufacturing technique has it´s own charcteristics also...For instance...Things like the orientation of the reinforcement tapes (running Block or Bias) can cause a ca.30% difference in drag for the same diameter PC, Though I applaude Andreas (mr) on his mathematical homework.. Unfortunately we don´t jump pc´s that are standardized rigid discs with the axis paralell with the windflow..That are from standardized construction techniques, using a standard of materials....also the rate of venting of air is highly variable between differing manu´s because of the assymetric construction techniques used today and sewing tolerances and even how you attach it to the bridle....As far as I know, I was the first person to market zp base PCs, I did it in response to the degradation of drag with f-111 over time and use, the videoed tests we (Lukas Knutson and I) did showed major differences in inflation times/descent rates for similar sized pcs weighted the same....so to sum it all up to the bottom line...it seems that a combination of a fish scale, extender bar and motorized vehicle is the way to go to set the standard...
just a thought,
space
ps, Hey Stane, we could benefit from your input here ;-)

[guest]

I want to make clear one point, after having read all previous posts after my original one.

The above table must be read in the sense that if you have a fully open and inflated P.C., then at a given delay you have that given drag force. It does not take into account at all the opening speed of each P.C., that is another completely different matter !. Of course, a smaller P.C. needs less air to inflate than a larger P.C., but a smaller P.C. could fall into the burble and a larger P.C. could make it out of the burble (simply, it is bigger). And, a larger P.C. pulls more ! ! !

And I agree with CPB when he says that the above drag force are (possibly) the drag force at the beginning of opening of container, what happens later is very complicated and very involved even to be sorted out theoretically. Still agree with CPB about the “relative” use of above table, also, helping for picking up the right size of P.C., and the above figures, I think, confirm the suggested P.C. sizes for given delays published by Mick Knutson and by Basic Research/Consolidated Rigging/Vertigo.

The comment from Craig. The formulas are there just to simply allow anyone to calculate his/her own terminal velocity and equation of motion for each own weight/shape in freefall. By the way, hyperbolic tangent (that could look difficult or strange) can be found on any modern pocket calculator, either under “tanh” itself or under pressing first “hyp” and then “tan”. In any case, if you are not very keen on mathematics, do not worry, the above table is “mathematically” true for a jumper weighing 80 kg, other weights do not change the numbers dramatically, so use the numbers as they are.

The comment from Tom. Consider the following. If you remember the assumptions I made about the validity of the table: “…they are actuall drag force values if, and only if, you have an infinitely rigid disc (i.e., steel plate) put in an airstream (with its axis perfectly parallel to the airstream itself) at that given speed and having no burble in front of it…”.

So, the only true thing is to compare drag force of different size P.C.’s at the same time, but only if they are completely open. Can you get the point ? If you do not have an infinitely rigid disc, those numbers are not valid at all. For example, a 28”, for sure, never exists “open” after 1 s if it falls into your burble, you know what I mean ? It is not for nothing that for BASE you use a longer bridle (9’ - 2.74 m) than when you skydive (7’ - 2.13 m). It is not for nothing that when you jump handheld, you “S” fold the P.C., so it inflates as early as possible. It is not for nothing that if you must do 1 s of freefall, you go for a 48” (being larger, once inflated, has 3 times the pull force than a 28”). The drag force values of the table start to approach the actual drag force values (even if they must be divided by the “diminishing factor”) only if you have something that resembles an fully open disc of that given diameter. Earlier than full inflation of P.C., the above numbers are useful only to play bingo. I think I read a post from you mentioning that when you did a low jump with P.C. stowed, it took longer to the P.C. to inflate than when you deploy from handheld: of course, when it is stowed and you get it out, after 0.5 s, probably it is not yet fully open, and so it is not yet a 48” P.C., imagine what can happen to a 28” after 0.5 s, probably the 28” can inflate faster (if it clears the burble), but has a very poor drag force, in the first seconds of freefall. I read posts from older BASE jumpers mentioning fatalities happened because of use of skydiving P.C.’s (28”÷32”) from low heights.

So, if we are talking of very small delay and very low airspeed, the number are very indicative, if what you have on air is far from being a “disc” of that diameter. If you have a large P.C. that you “S” folded so to have it open immediately, then, when it is open and inflated, the drag force values could be close to truth.

You get drag forces as per above (or close to) only if you get a perfectly open and inflated P.C., earlier than that stage, you probably get 0 kg of drag force, especially if the P.C. falls in the burble of your body, for example, if you used a skydiving bridle (7’ - 2.13 m) and if you failed to launch your P.C. properly off your side, you do not get an open and inflated P.C., and so, you get 0 kg of drag force.

Think just the following. Try to get a perfectly inflated 28” after 1 s of freefall on a BASE jump: no chance, if it falls into your burble ! You would get a 28” open after 1 s only if you put into it a circular shaped steel spring (like the spring used in the plastic sunshades you use in your car). That’s why you use, for a 1 s delay a 48” and for a 3 s delay a 42”, folded both in “S” fashion, if you handheld, or, if you stow it, at least folded with the mushroom technique to get it open as soon as possible. First, you must get your P.C. fully open and inflated, and then you talk about drag force.

Unless you have a completely open and inflated P.C., it has no sense talking about any drag force, so, the drag forces reported for very small delays, unless you have a circular spring loaded P.C. fully open and inflated and very far from your burble, are just mathematical numbers, with no meaning.

One thing is for sure: if you have both a 48” fully open and inflated and a 28” fully open and inflated, at any given delay for which such a condition holds true, it is true that drag force of 48” is 3 times the drag force of 28”.

So, if you have a fully open and inflated P.C., look at the table and read the relevant drag force (still the “diminishing factor” remaing pending…); if you have a P.C. still not inflated, you do not have a P.C., above your head you have only scrap fabric hanging around your burble.

Of course, experimental results would help the thread, probably generating a “diminishing factor” of 2 or so; still I think it holds the truth of above table if read in a “relative” way.

Blue skies

Andrea Checchia

[guest]

Many years ago I did some really fun pilot chute tests with the guys from the Relative Workshop in Deland. It was around 1979 or 80 and the emphasis was on less pack volumne and ease of manufacture. One of the most curious things we discovered were some pilot chutes made entirely of 1.1 rip-stop that had patterns of holes cut in the lower surface to allow for inflation. There were several different shapes including round, square and octagonal. These different itterations were amazingly strong pulling, far stronger than comparable ones with mesh. They were really tiny and easy to make. Their fatal flaw was they wouldn't inflate worth a damn. Once they did though watch out... the snatch force was higher than anything we'd previously used. Many of the experimental models never made it further than out the back of a pick-up truck.

There were a lot of figures taken with the fish scale but I wouldn't think any have survived as a retrieveable document.