Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!zaphod.mps.ohio-state.edu!think.com!spool.mu.edu!uunet!lhdsy1!yzarn From: yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) Newsgroups: rec.skydiving Subject: Re: Graph of osc'n due to high windforce Message-ID: <997@lhdsy1.chevron.com> Date: 25 Jun 91 18:00:59 GMT Article-I.D.: lhdsy1.997 References: <1991Jun21.145113.6028@Stardent.COM> <994@lhdsy1.chevron.com> <1991Jun24.154627.285@Stardent.COM> Organization: Chevron Oil Field Research, La Habra, CA. Lines: 66 In article <1991Jun24.154627.285@Stardent.COM> joep@Stardent.COM (Joe Peterson) writes: > >> Nooooo. I said one down (assuming g not changing with altitude, and >> remember, Earth is attracting us, so g is toward the center of the >> Earth) and 0.023 (rounded) up, which would mean his acceleration is less >> than one. >> Look at it this way, since we agree that the skydiver's terminal >> velocity is slowing down as he/she gets lower in the atmosphere, how >> could his/her acceleration be greater than g? I don' get it. > >Well, what I was saying was that the skydiver would FEEL a force larger >than 1.0 g. Just like when you open your parachute, you feel a force >much larger than 1.0. If a person is falling at a constant velocity in >a constant gravitational field, he/she will feel the same force as when >lying on the ground. On the ground, there is the force due to gravity >acting downward and the "normal" force acting upward (the support of >a solid surface). These are equal and opposing forces. In freefall, >the force of gravity downward would be equally opposed by the force >of friction upward (once terminal velocity is reached) if we assume >no slowing down as the atmosphere gets more dense. Now, if the person >is slowing down, the force due to friction is constantly higher than >that due to gravity. This is an additional force felt by the skydiver >pushing upward, making him feel heavier. Therefore, he is "pulling >more than 1.0 g's." It is the same as being in an elevator and starting >to go up. The net force upward is greater than 1.0 g's for a second or >so. Does this make sense? I knew that at some point the elevator analogy would be brought up! The reason why when a skydiver opens his/her parachute, he/she feels it and *it is greater* than 1 g only because the person goes from 100mph (say) to 10-15' per second (3 mph???) in a *short* time and since acceleration is measured in unit of length/second/second then you obtain a fast deceleration say (100 - 3)*1.6*1000/3600 [mph*km/miles*meters/km*seconds/hour]= 43 meters/second = velocity loss. Say the parachute takes 2 seconds to open 43/2 = 21.5 m/s/s Since g = 9.81 m/s/s, the deceleration is 21.5/9.81 [m/s/s*g/(m/s/s)] = 2.2 g deceleration. Yes, it is above one, but only because the amount of time it took to slow down was short. And this is the net difference. Now, I agree that the net upward force on a skydiver in freefall in an atmosphere (Earth!) will be greater than 1 g, in the case discussed previously (a few exchanges ago), it would be around 1.025 g? *But* 1 g would be cancelled because of the net downward gravitational attraction. So the skydiver would *feel* 0.025 g which is unfeelable by human standards! Using your analogy, when a person is in an elevator, he/she does not feel 1+ g but the acceleration of the elevator. When I am standing up, Earth is pulling down 1 g, the surface I am standing on compensates (so I don't sink into the floor) and I will argue that I don't feel the floor pushing me up (to avoid sinking into it!) I can also say that I do not feel my weight either because my muscles are used to and can compensate (for now!) for the gravitational pull. When you do analyze physical problems, you analyze all forces that apply but the resultant (or summation of all the forces) is what will determine what happens to the system under study. A person at rest on the ground does not pull one g! Yet this is what you argue a skydiver feels in freefall. I disagree. I look forward to your next entry, this is stimulating. E.F.S. -- Philip Yzarn de Louraille Internet: yzarn@chevron.com Research Support Division Unix & Open Systems Chevron Information & Technology Co. Tel: (213) 694-9232 P.O. Box 446, La Habra, CA 90633-0446 Fax: (213) 694-7709