remko

> He had mentioned that he had good openings when the plane was in a dive. Hmmm... I don't recall saying that. The plane was never in a descent (dive is a bit much). > And I do believe that > the dive will have more of an impact if you are last out versus first out. Makes no difference, unless the plane is accelerating. > I know I for one was trying to make a choice on every exit. Do I go for a > quick opening, not as hard, and then work into a crazy dizzying spiral to get > to down to the formation already built? I generally was taking 4-5 seconds > and spiraling some. I had the same dilemma and choose for soft openings and more work to get down. All though you really don't have a choice, you must jump as briefed. Exiting with 500'/min & 3.0 sec delay seems like a good trade-off. > It seems that one of the things we need is to train jump pilots on exit trims > for CRW exits. There are a few things a pilot can do about exits. Slow down for one, using flaps. But there is a balance between safety and comfort. A slow flying plane (close to stall speed) is a dangerous one. Some pilots choose for comfort and a little less safety some for more safety and a little less comfort. Jumpers can help A LOT by staying in the front of the plane as long as possible. If the pilot knows the jumpers are disciplined and do this consistently he can slow down a bit more without jeopardizing safety. Climb. As shown above this will decrease deployment speed a bit (like throwing up a ball). furthermore the increased airflow over the wing by the prop-wash will allow him to slow down even more, using landing flaps requires even more power. But I think this is not what we want on a big-way jump-run. > A good friend has left CRW because of hard openings. Maybe use a RW triathlon and re-line it for CReW... > It's > got to be an issue for more than just the two of us. Establishing the ideal > exit trim would promote safety in the sport. So, stress on staying/moving to the front of the plane on exit! -- Everything you know is wrong. But some of it is a useful first approximation.

When the weather sucks there is time to play with some numbers. At the WR I figured why not drop in a descent, since the formation is sinking with about 1300'/min. Hard openings was the answer. I couldn't believe a little descent (1000'/min = 5 kts) would make that much of a difference but the answer is not as easy as I thought. Horizontally the jumper decelerates due to drag. Vertically the jumper accelerates due to gravity minus the drag. Drag depends on speed. So I wrote this little snippet of C to figure out deployment speeds. Depending a little bit on the numbers you feed it (jumpers mass, dragcoefficient and altitude) I came up with the following: Plane flying at 13.000 feet with 78 kts, standard atmosphere. Level: At 3.0 sec delay the deployment speed is 75.9 kst. The lowest speed is found at 1.5 sec: 70.3 kts. 500'/min: At 3.0 sec delay the deployment speed is 78.4 kts. This equals to a delay of 3.4 sec flying level. The lowest speed is found at 1.3 sec: 72.0 kts. 1000'/min: At 3.0 sec delay the deployment speed is 80.8 This equals to a delay of 3.8 sec flying level. he lowest speed is found at 1.1 sec: 73.5 kts. 500'/min climb: At 3.0 sec delay the deployment speed is 73. kts. The lowest speed is found at 1.6 sec: 68.4 kts. 1000'/min climb: At 3.0 sec delay the deployment speed is 70.5 The lowest speed is found at 1.7 sec: 66.4 kts. #include #include main() { const double Na, R, g, mmt, p0, T0, y, kts, ft; double Cd, S, m, h, a, ay, vx, vy, v, t, dt,\ W, M, T, p, rho, dx, dy; Na = 6.0221E+23; /* constante van Avogadro */ R = 8.3143; /* gasconstante */ g = -9.8; /* gravitatie constante */ mmt = 4.80793E-26; /* molecuulmassa troposfeer */ p0 = 101325.0; /* standaard druk */ T0 = 273.15 + 15.0; /* standaard temperetuur */ y = -.0065; /* temperatuur gradient */ kts = 3.6 / 1.852; /* meter/seconde => knopen */ ft = 1 / .3048; /* meters => voeten */ Cd = 1.2; /* weerstands coefficient */ S = .6; /* frontaal oppervlak */ m = 85.0; /* massa lichaam */ h = 4000.0; /* afpringhoogte in meters */ dt = .1; /* delta tijd */ v = 40.0; /* totale snelheid */ vy = -0.0; /* vertikale snelheid */ vx = sqrt( v * v - vy * vy ); /* horizontale snelheid */ W = (Cd * .5 * S) / m; /* weerstands variabele */ M = Na * mmt; /* molaire massa troposfeer */ for( t = 0.0; t

Also the brake setting matters. I have a lightning with a deep brake setting that opens considerably softer/slower than one with a standard brake setting. Greetings, Remko -- Everything you know is wrong. But some of it is a useful first approximation.

The *.jpg worked fine, --- Off topic --- but that's like giftwrapping a postcard in a wooden box. cvs Stands for comma separated value and should be platform independent except that microsoft uses a Carriage Return + Line Feed in stead of a simple New Line, so `mac' would be the best choice. Plain text would be ideal to communicate in most cases unfortunately microsoft likes to make up their own little secret languages every year to tempt you to buy or even steel their newest software. --- End of off topic --- Remko -- Everything you know is wrong. But some of it is a useful first approximation.

Can you send a *.txt or *.csv maybe? Cheers, Remko -- Everything you know is wrong. But some of it is a useful first approximation.

[Waking from hybernation] Sent: Wed, 06 Mar 2002 03:38:07 -0700 Subject: How does a canopy turn? When I asked myself this question a while ago it kept me puzzled and awake at night for quite some time and it doesn't surprise me at all that after so many posts only Paul has touched the essence of the solution to the canopy turning problem by mentioning the pendulum effect. The BIG difference between an airplane and a canopy is the position of the aerodynamic center and the center of gravity. In an airplane these two are close together. Moving the aileron itself will not turn the plane but only roll it, rolling the lift vector with it creating a sideways pointing component that will turn the plane. During roll-in, the downward deflected aileron on the outside of the turn will create more lift during the roll-in thus, as mentioned before, create more induced drag. The upward deflected aileron on the inside of the turn will decrease lift, thus less induced drag. this will yaw the plane to the opposite side (outside) of the turn. This unwanted side effect is called adverse yaw and is compensated for with rudder into the turn during the roll-in and out of the turn during roll-out. Note that aileron is used only to roll-in and roll-out of the turn and not during the turn and that the resulting decreased vertical lift component will make the plane decent if not countered. Under a canopy the center of gravity and the aerodynamic center are some 10 feet apart, depending on the line lenght. Now, pulling a toggle or rear riser down will increase lift but the weight of the jumper on that extreme long arm will prevent it from rolling. The increased drag however will yaw it, swinging the jumper to the outside (in fact, it's momentum will want to keep it going in a straight line while the canopy turns), rolling the wing and with it the lift vector, creating the sideways force. Note that the "aileron"(toggle) is used throughout the turn and that the turn is a combination of "adverse" yaw caused by increased drag and rolled lift vector caused by the pendulum effect. Also the decreased vertical component will increase the decent rate. If continued this will develop into a spiral dive which is also confusingly called a stall turn. Remko -- skydive long and prosperous PS: A toggle/rear-riser increases camber as well as angle of attac, a front-riser only increases camber.

Part of packing is inspecting the canopy. Since it gets jerked at a lot doing CReW I leave it out after the last jump of the weekend. This has two advantages. First it enables me to drink beer sooner, secondly I take some extra time during the week to give it an extra look. I check for little burn holes in the fabric and mark them so I can differentiate the new ones from the old ones. I check for loose stitches, dents/loose/sharp edges on slider/container grommets, corrosion on quick links (especially under the lines) and the lines... It's easy to spot a damaged line, but I am also interested in checking the length. I want to know how much my canopy is out of trim and how much is much or too much. This is where the mistery starts. I understand that, for a lightning anyway, the lines of a canopy is like paint for a car. You can order all kinds of lengths and trims. The subject is more complicated than I first thought due to the cascades (assuming for simplicity that ALL lines are cascaded). Is it true that all A-lines should be equally long? In other words that the canopy shape is circular, left to right, when the connector links are held together? And that the difference in A/B/C/D/Steering lines is linear? In other words that the under side, front to aft, of the canopy is straight? And how much difference is acceptable, 1 inch, 2, 3? I heard someone say that A lines are in principle cut as a multitude of feet, true? But lines shrink A LOT as a result of produced heat by for example slider movement. How much is "a lot"? 1 foot, 2? Remko -- Skydive long and prosperous.