Change Forum... Civil Aviation Travel, Polls & Prefs Tech/Ops Aviation Hobby Aviation Photography Photography Feedback Trip Reports Military Av & Space Non-Aviation Site Related LIVE Chat My Starred Topics | Profile | New Topic | Forum Index | Help | Search
 I Need Some Serious Algebra Help
 Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 20Posted Fri Dec 14 2001 09:05:55 UTC (13 years 3 months 2 weeks 5 days 15 hours ago) and read 1419 times:

 Here’s the deal: Rather than trying to construct a wind tunnel and calculate various numbers to find the coefficient of lift of an airfoil, I thought of simply dropping a model airplane from some known height of Y. The force of lift will make it follow some curve away from a point on the ground at a distance of X. The picture below shows what I’m trying to get at and all of the formulas I’ve used. The problem is I can’t get the formula to be without the terms of time and velocity. I know I’m missing something because every time I do it, my coeff has units! Where g = 9.80m/s squared, rho (density) = 1.22 kg/m3, uF (coeff of friction) = assumed to be 1.0 [I don't know if I am dealing with coeff of friction or coeff of drag] To check my answer for realism, the weight (mg) must equal the lift force (Fx) with speed (vy) of 3m/s, it never does. What am I doing wrong?
 The meaning of life is curiosity; we were put on this planet to explore opportunities.
 Airways1 From United Kingdom, joined Jul 1999, 562 posts, RR: 0 Reply 1, posted Fri Dec 14 2001 14:43:29 UTC (13 years 3 months 2 weeks 5 days 10 hours ago) and read 1370 times:

 I'm sorry but the situation is a lot more complicated than you have made out with your maths. First mistake, the acceleration that you have called a is actually proportional to vy-squared, not vx-squared. Why is this? Well you are considering acceleration in the X direction, so it should be equal to Fx/m, and Fx as you stated in your first equation is equal to vy-squared. Second mistake, those equations you have quoted for the value of Y assume a constant acceleration, and this is not the case since you have a term proportional to vy-squared in your expression for Fy. If you want to calculate it properly you will need to use calculus. Apart from all that, you assume that your model plane will remain at a constant angle during its whole descent, which is unlikely, plus you have neglected drag in the X direction, which might be important in practice. I suggest you go back to the wind tunnel idea where everything is operating under steady conditions.
 AgnusBymaster From United States of America, joined Feb 2001, 652 posts, RR: 0 Reply 2, posted Fri Dec 14 2001 17:11:11 UTC (13 years 3 months 2 weeks 5 days 7 hours ago) and read 1367 times:

 I have to admit, this is not what I expected when I opened this thread!
 Olympic A-340 From United States of America, joined Apr 2000, 780 posts, RR: 10 Reply 3, posted Fri Dec 14 2001 22:40:37 UTC (13 years 3 months 2 weeks 5 days 2 hours ago) and read 1342 times:

 American_4275 From United States of America, joined Aug 1999, 1076 posts, RR: 0 Reply 4, posted Fri Dec 14 2001 22:42:20 UTC (13 years 3 months 2 weeks 5 days 2 hours ago) and read 1339 times:

 Agnusbymaster, I agree 100%. Lehpron, Good luck dude that looks horrid!
 777236ER From , joined Dec 1969, posts, RR: Reply 5, posted Fri Dec 14 2001 22:53:02 UTC (13 years 3 months 2 weeks 5 days 2 hours ago) and read 1333 times:

 Argh, you've messed up quite a bit there. Airways1 spotted them all, i think. my coeff has units! Lol. I love it when that happens. I've fucked up SO much in the past cos coeffs have units. Just today, a coeff of restitution was 5 (!) and had units Kg/m!!! The particles were hitting each other, there was a completly elastic collision, but the system ended up with more mass than it started with! ARGH! In my expierence, theoretical windtunnel formulas look good, but there are too many variables for them to be too accurate in real life. Still working on the SC replacement? I'd like to see the updated version! How's the sonic-boom-reducing lamina flow going? Regards 777236ER
 Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 20 Reply 6, posted Tue Dec 18 2001 06:35:59 UTC (13 years 3 months 2 weeks 1 day 18 hours ago) and read 1297 times:

 Airways1: ...the acceleration that you have called a is actually proportional to vy-squared, not vx-squared. Why is this Think of it this way: Lift depends on the velocity squared, assuming everything else is somewhat constant. Since velocity is governed by the fact than the force of gravity accelerates the planes down, hence Vy. ...you have quoted for the value of Y assume a constant acceleration... It isn't constant, that's why I put Fy=mg-(.5)(rho)A(Vy^2)Cf. As the plane falls gravity pulls it down and drag friction keeps it from accelerating too fast, hence terminal velocity. On the other hand, I figure that since I am droping this thing in a room, the variance is somewhat negligible. ...you assume that your model plane will remain at a constant angle during its whole descent, which is unlikely Why? The center of lift and the center of gravity are at the same point and with no elevator flaps or moments changing the AoA, it really cannot vary. ...plus you have neglected drag in the X direction, which might be important in practice. Sorry, I forgot this! Yeah, no it's a whole lot easier to figure out, huh?     777236ER: Still working on the SC replacement? I'd like to see the updated version! How's the sonic-boom-reducing lamina flow going? Yes I'm still trying to tweek it so the area rule works as well as the current problem of have induced drag at one end of the wing; aside from finals, it's all good. I haven't gone to my site and updated anything since I've been busy. Me and my engineering proffesor are going to be working on the engine phase of Project: Lehpron, I plan to introduce this as my entry into the 2002 Collegiate Inventor Competition this coming year in Ohio. Lehpron is my brainchild, she's supposed to be about the size of NASA's Hyper-X, except she's supposed to takeoff on her own, it's a challege that represents my primary interest in commercial airliners: the advent of Hypersonic airliners. I can't help it, I like hypersonic more than supersonic, do you blame me? That's probably why I'm getting behind, don't worry you guys, the "cm_8Wa" will be inaugurated by mid-January and will be revealed.   lehpron
 The meaning of life is curiosity; we were put on this planet to explore opportunities.
 Airways1 From United Kingdom, joined Jul 1999, 562 posts, RR: 0 Reply 7, posted Tue Dec 18 2001 14:37:59 UTC (13 years 3 months 2 weeks 1 day 10 hours ago) and read 1285 times:

 lephron Your acceleration, a, is in the X direction, therefore it should be equal to Fx/m, not Fy/m as you have stated. Secondly, I see you have the term proportional to vy-squared in your expression for Fy. However, I shall say again, the equation you have quoted for Y assumes a constant acceleration. In other words, your equation for Y is not compatible with your equation for Fy. As for your aerofoil remaining at a constant angle, then I agree with you it should be OK if the centre of lift and centre of mass are coincident, but that assumes the position of the centre of lift is independent of the velocity.
 777236ER From , joined Dec 1969, posts, RR: Reply 8, posted Tue Dec 18 2001 15:21:00 UTC (13 years 3 months 2 weeks 1 day 9 hours ago) and read 1280 times:

 Bah, hypersonic isn't flying! It's just blasting through the air, knocking everything in front of it out of the way. Seriuosly though, if you've come up with something you regard as being unique and good, you should patent it. It's very expensive, but it's worth it. Trust me. I had a design that i have produced stolen by a collegue once, and without a patent, you have almost no legal ground to show that you have the rights to the design, and not him. Regards.
 Delta-flyer From United States of America, joined Jul 2001, 2676 posts, RR: 6 Reply 9, posted Wed Dec 19 2001 03:04:46 UTC (13 years 3 months 2 weeks 21 hours ago) and read 1264 times:

 As the plane falls gravity pulls it down and drag friction keeps it from accelerating too fast, hence terminal velocity. Lehpron, if you are assuming that terminal velocity is reached, then the trajectory of the plane will be a straight line (ignoring the initial transient from zero to terminal velocity). The only way for this experiment to make sense is if the plane could be made to glide at a constant angle of attack. Then, by conducting tests at different aoa's, it may be possible to map lift and drag coefficients, but I can't imagine any way of holding aoa to a given value. At least not without little pilots on board! That's what wind tunnels do, which is where we started.   Cheers Pete
 Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 20 Reply 10, posted Wed Dec 19 2001 04:43:11 UTC (13 years 3 months 2 weeks 20 hours ago) and read 1256 times:

 Airways1:...the equation you have quoted for Y assumes a constant acceleration. In other words, your equation for Y is not compatible with your equation for Fy. If this is the case, what should/could it be? Delta-flyer:...then the trajectory of the plane will be a straight line... Perhaps stating the terminal velocity thing was misleading, this model will never reach a terminal velocity in a 2.4m room! I figured the calculations would be simpler if the drag due to falling were ignored because it might not have such a big effect on force considering it wouldn't be going all that fast. But it is a paper-airplane, in fact it is an 80%-approx to Lehpron. I can't find much info online about flying wedges so I thought this type of analysis would help.
 The meaning of life is curiosity; we were put on this planet to explore opportunities.
 Airways1 From United Kingdom, joined Jul 1999, 562 posts, RR: 0 Reply 11, posted Wed Dec 19 2001 11:35:15 UTC (13 years 3 months 2 weeks 13 hours ago) and read 1249 times:

 Lephron It's a bit difficult to show you what the equation should be without being able to display the correct mathematical symbols. Let's just say, use Newton's 2nd law in the form F=m(d²y/dt²) and put in your value for Fy, remembering that vy=dy/dt. Then you will be left with a differential equation that you need to solve for y.
 Top Of Page Change Forum... Civil Aviation Travel, Polls & Prefs Tech/Ops Aviation Hobby Aviation Photography Photography Feedback Trip Reports Military Av & Space Non-Aviation Site Related LIVE Chat Forum Index

This topic is archived and can not be replied to any more.

Printer friendly format

 Similar topics: More similar topics...
I Need Some Serious Help posted Tue Apr 19 2005 08:49:12 by Tbar220
I Need Some Mariachi Music...Help! posted Sun Apr 30 2006 20:51:49 by DeltaGator
Need Some Help From Zee Germans... posted Fri Aug 4 2006 12:38:47 by Thom@s
Need Some Help With Some States In The U.S posted Sun May 21 2006 22:42:07 by Christeljs
I Need Some Help On Spain! posted Mon Mar 13 2006 05:27:25 by COEWR777
Need Some Help With My Budget posted Wed Dec 21 2005 21:04:30 by Tbar220
Need Some Help With RSS/Sage posted Thu Mar 3 2005 17:50:58 by DesertJets
Need Some Help About A Cartoon Sitcom! posted Fri Nov 26 2004 08:53:30 by OYRJA
I Need Some Help Finding A CD... posted Wed Sep 8 2004 03:44:53 by Iamcanadian
Need Some Plannning Help posted Wed May 12 2004 19:14:40 by United4EverDEN