Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 19 Posted (14 years 2 months 4 days 7 hours ago) and read 1870 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 (14 years 2 months 4 days 1 hour ago) and read 1821 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 (14 years 2 months 3 days 23 hours ago) and read 1818 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: 9
Reply 3, posted (14 years 2 months 3 days 17 hours ago) and read 1793 times:

American_4275 From United States of America, joined Aug 1999, 1076 posts, RR: 0
Reply 4, posted (14 years 2 months 3 days 17 hours ago) and read 1790 times:

777236ER From , joined Dec 1969, posts, RR:
Reply 5, posted (14 years 2 months 3 days 17 hours ago) and read 1784 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?

Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 19
Reply 6, posted (14 years 2 months 9 hours ago) and read 1748 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 (14 years 2 months 1 hour ago) and read 1736 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 (14 years 2 months ago) and read 1731 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.

Delta-flyer From United States of America, joined Jul 2001, 2679 posts, RR: 6
Reply 9, posted (14 years 1 month 4 weeks 1 day 13 hours ago) and read 1715 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.

Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 19
Reply 10, posted (14 years 1 month 4 weeks 1 day 11 hours ago) and read 1707 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.