dropdeded 0 #1 February 9, 2003 I start a Physiology class tomorrow,wtf is physiology? Its for an Emergency Services course. dropdeded------------------------------------------ The Dude Abides. - Quote Share this post Link to post Share on other sites
LouDiamond 1 #2 February 9, 2003 Main Entry: phys·i·ol·o·gy Pronunciation: "fi-zE-'ä-l&-jE Function: noun Etymology: Latin physiologia natural science, from Greek, from physi- + -logia -logy Date: 1597 1 : a branch of biology that deals with the functions and activities of life or of living matter (as organs, tissues, or cells) and of the physical and chemical phenomena involved; compare ANATOMY 2 : the organic processes and phenomena of an organism or any of its parts or of a particular bodily process - phys·i·ol·o·gist /-jist/ noun More importantly, this is very applicable to high altitude skydiving and understanding how the human body is affected at altitude and the lack of O2 in the blood stream and how Nitrogen plays in to the equation as well. So now you have a good topic to write your paper on, "The physiological effects of high altitude skydiving on the human body by Joe skydiver(insert your name here) "It's just skydiving..additional drama is not required" Some people dream about flying, I live my dream SKYMONKEY PUBLISHING Quote Share this post Link to post Share on other sites
ECVZZ 0 #3 February 9, 2003 Quotewtf is physiology? In a sentence: A pain in the ass! G. Quote Share this post Link to post Share on other sites
dropdeded 0 #4 February 9, 2003 More importantly, this is very applicable to high altitude skydiving and understanding how the human body is affected at altitude and the lack of O2 in the blood stream and how Nitrogen plays in to the equation as well. So now you have a good topic to write your paper on, "The physiological effects of high altitude skydiving on the human body by Joe skydiver(insert your name here __________________________________________________ Dude,that is a most excellent ideaThanks dropdeded ------------------------------------------ The Dude Abides. - Quote Share this post Link to post Share on other sites
smiles 0 #5 February 9, 2003 Ya think you'll be learning how to figure out gravitational forces?? http://www.adtdl.army.mil/cgi-bin/atdl.dll/fm/3-04.301/ch4.htm#fig4-3 Long thread at chat@cspa.ca (canadian sport parachute association) started with this question: A couple of weeks ago one of my toggles became unstowed during deployment. I was flying a Sabre2 190 loaded at about 1.21, when my brake line became unstowed. Does anybody have any idea what kind of G forces a canopy pilot experiences when he performs a full arm extension toggle turn that induces a spiral or goes into a spiral because a brake unstows during a deployment? Holger Marten With responses such as this: Everything I am about to say is based on a guess only, and no real data. I think the max G force you could experience under a canopy is less than two. Why do I say that? I know, 100%, that I have experienced 2 Gs. An airplane, any airplane, when flown at a 60 degree bank angle, such that it is neither climbing nor descending, experiences exactly 2 G. It does not matter the type of plane, it does not matter the speed at which this turn is executed at, all that matters is the bank angle and the fact that you are remaining at a constant altitude. Having done this dozens of times, I have a reasonable idea of what 2 Gs feel like. I have never experienced this amount of G under a canopy. I jump a Stiletto 120 loaded to about 1.54 lbs / square foot. I admit that I don't push it too hard, but I don't baby it either. Are there any electronics geeks out there? How hard would it be to couple a small G sensor up to a recording device? If the CSPA wants something new to direct their energies towards, how about coordinating and sponsoring this type of research? It benefits everyone. Just a thought. Donald Gravelle I suppose that some are dying to experience super high g-force turns under canopy.......unsure if any studies done with monkeys?...inertia and g-force from high speed twisting mal/ human tolerance on g-forces.............. Smiles Quote Share this post Link to post Share on other sites
Mad47 0 #6 February 9, 2003 When you get to 13.5K feet and realize that you terrible want to take a leak, it doesn't indicate that you're scared. It's physiology Quote Share this post Link to post Share on other sites
smiles 0 #7 February 10, 2003 now we have it...(small world) thanks to a skydiving physics teacher: André Lemaire ---The G forces are related in a spinning malfunction in the following way. 1) a mass which is spinning or changing its direction continuously experiences a centripetal (toward the center of rotation) acceleration (change of the velocity vector during a change of time) 2) the following formulas come from the study of the circular motion and they are : a) a=(VxV)/R or (V square divided by R) b) a = (4 x pi x pi x R)/(T x T) or (4 x pi square x R) divided by T square c) a = (2 x pi x V)/T or (2 x pi x V) divided by T WHERE : a = acceleration (in meter/second square) V = velocity or speed ( in meter/second)) R = radius of rotation (in meter) T = period of rotation (time for one rotation) (in second per turn) 3) Now applying the Newton's second law of motion: F = M x a (force = mass x acceleration) Force in Newton (1 Newton is about 1/4 of lbs), mass in kilogram ( 1 kilogram is about 2.2 lbs), acceleration in meter/second square. Then if you want to get the force involved when spinning: You just multiply the acceleration calculated with one of the 3 formulas above (item 2) by the spinning mass (M) (make sure you respect the units) EXAMPLE: A person of 70 kg (154 lbs) is spinning under a canopy at the rate of 2 turns per second (quite fast) and the radius (horizontal) of rotation is 6 meters (20 ft) . The radius has to be taken from the vertical axis of rotation (not necessarely the parachute since there is probably a spiral made by the canopy itself) to the center of gravity of the spinning mass (ie. the belly button of the jumper). Calculations: I will choose the formula b) since I have the period of rotation (2 turns/s) and the horizontal radius of the rotation (6 meters) . then a = (4 x 3.14 x 3.14 x 6)/(2 x 2) = 59.16 m/s square (1 G = about 10 m/s square) here in this case you have 59.16 / 10 = about 6 Gs (a lot) meaning 6 times your weight If you want to put that in terms of forces. F = M x a then F = 70 kg x 59.16 m/s square= 4141 Newton or about 1000 lbs (a lot) Conclusion : spinning a the rate of 2 turns per second is quite fast, if the horizontal radius(which can be more than the length of the lines if there is an actual spiral path) is 6 meters, then you will experience about a 1000 lbs of force on the lines. (remember, about 6 times your actual weight of 154 lbs in this example) If this is not clear, let me know. As a physics teacher, this is my job to spread the knowledge of the physics laws and make them clear. Physically yours André Lemaire now I have learned the difference between physics & physiology!!! Smiles Quote Share this post Link to post Share on other sites
Push 0 #8 February 10, 2003 Just a quick comment. The force is slightly higher than that because this is a spiral and the calculations are for a circular path. The gravitational force does add a certain component, though I imagine it's fairly difficult to ascertain exactly how much. On the other hand, you are always told how fast your canopy is diving when it's spinning, and the dive happens entirely due to gravity, so the extra G can be pretty significant. So what actually happens is that the aerodynamics of the canopy (Lift) are causing the circular component of the spiral, and gravity is causing the downward motion. The net force is somewhere in between. All that crap means that it's fairly difficult to design a device that will measure the Gs exactly, because this force is in no particular direction and depends on many different things. I'm sure it's possible though. You can get a good picture from a tension scale on the risers and a gyroscope, for sure. Point is, it's nothing to joke about. Somebody in Incidents I believe posted about a spinner they had that gave them tunnel vision. Cue sig line. -- Toggle Whippin' Yahoo Skydiving is easy. All you have to do is relax while plummetting at 120 mph from 10,000' with nothing but some nylon and webbing to save you. Quote Share this post Link to post Share on other sites
masher 1 #9 February 10, 2003 The maximum that accel'n due to gravity will add is 1 g. The inertial forces that you experience are made of two components: the circular motion of yourslef under the canopy, and the pull of gravity. Now, measuring these two components is fairly difficult, as you need two accelerometers that are (at all times) perpendicular and parallel to the ground (to measure centripital and gravitational accel'n). Measuring the force on you body is a bit easier. Taking an idea frp, Push, you can mount something like a fish scale, with a thingo that records the highest weight (like a tag that goes down, but not up) and attach a weight to the end. This is mounted vertically on your leg, so the the weight is pulled down to your feet. After you deploy (you may have to reset the scale, I don't know about g-forces on deployment), just put yourself into a spin, making sure to keep you legs pointing straight. After landing (softly) look at the max weight. That will tell you how many g's you experienced. ie. if you put a 100g mass on the spring, and the max weight is 200g, then you experienced 2 g's.-- Arching is overrated - Marlies Quote Share this post Link to post Share on other sites
dropdeded 0 #10 February 10, 2003 After reading all the replies,my next thought is,holy fuckin crap.I thought it had to do with cells and shit.dropdeded ------------------------------------------ The Dude Abides. - Quote Share this post Link to post Share on other sites
masher 1 #11 February 10, 2003 Thats true for circular motion in the absence of a gravitational field. A spinning parachutist is an example of a conical pendulum. More exactly, a conical pendulum with a constant driving force. For a normal conical pendulum, the only driving force is that of gravity. For a small body of mass, m, suspended from a string of length L, the body revolves in a horizontal circle of radius r. The angle, theta, describes the angle between the vertical and the string. Do the maths, and the period of revolution is 2*pi*sqrt((L*cos(theta))/g) This problem is 'easy', as the only acceleration is in the horizontal plane, and the source of the acceleration is gravity. With a parachutist you have the additional problem of acceleration in the vertical direction, as well as a driven osccilator; the parachute gives you forward speed, as well as gravity. That makes things hard.-- Arching is overrated - Marlies Quote Share this post Link to post Share on other sites
masher 1 #12 February 10, 2003 The theory is too hard for me to figure out at the moment (the book that I need is at home). To do it experimentally, you need a gyroscope setup and 3 accelerometers. You can integrate the z data twice to get your displacement (altitude loss). the z data should give a zero accel'n. The reason for this is the same as that behind freefall velocities. The only difference is that under canopy you have a larger surface area. So, once you start your turn, after the initial accel'n period, you should stay at a constant velocity. So, the path that you describe in that air is that of a driven conical pendulum falling at a constant velocity, and because you're falling at a constant velocity, we don't need to worry about it. (Eintein said so) So we only need to worry about accel'n in the horizontal plane. Convert the x and y accel'n to a radial accel'n. and you have the g forces that you are under. . There's no acceleration in the vertical (once you reach equilibrium). All of the acceleration is contained in the horizontal plane. Sorry about that, but I actually did some thinking before I wrote this post.-- Arching is overrated - Marlies Quote Share this post Link to post Share on other sites
masher 1 #13 February 10, 2003 I had another brain fart. Here we go. This should work. The velocity for a conical pendulum is given by v=sqrt(r*g*tan(theta)). The velocity for a driven conical pendulum is given by v=sqrt(r*g*tan(theta)) + x, where x is the constant velocity of the parachute. Centripital acceleration is given by a=v^2/r, so after all the maths, the acceleration that you experience is given by the equation: a = g*tan(theta) + 2*x*sqrt(r*g*tan(theta)) + (x^2/r) a = centrepital acceleration g = acceleration due to gravity theta = angle between the vertical and angle of revolution x = forward speed of you canopy-- Arching is overrated - Marlies Quote Share this post Link to post Share on other sites
LouDiamond 1 #14 February 10, 2003 Quote After reading all the replies,my next thought is,holy fuckin crap.I thought it had to do with cells and shit.dropdeded Yeah, I was thinking the same thing. How did we go from physiology to physics? This smells of a classic hi jackIn all seriousness, for the physic freaks, I think cobaltDan at Atair could probably give you some real data from actual testing."It's just skydiving..additional drama is not required" Some people dream about flying, I live my dream SKYMONKEY PUBLISHING Quote Share this post Link to post Share on other sites
masher 1 #15 February 11, 2003 QuoteHow did we go from physiology to physics? You need the physics to figure out the physiological effects. Plus, I didn't want to do any work at uni.-- Arching is overrated - Marlies Quote Share this post Link to post Share on other sites
