aonsquared

platypii

It wasn't constant airspeed nor constant glide. The flight went through a few phases and angles of attack, but as much as possible, the same body configuration.

Then your hypothesis is already incorrect. Why should L/D ratio remain constant if you change angle of attack?

Roughly constant body configuration should result in roughly constant profile drag though. Is that what you're calculating?

platypii

Hypothesis: once up to speed, your L/D should be roughly constant throughout the flight. Here is an example of such a flight.

You don't need yuri's equations for that. Any glider pilot knows that on a constant airspeed and constant glide, your L/D ratio is roughly constant. What's new?

Also, how was the green line calculated?

There are 2 definitions of "proof". He's talking about mathematical proof, which only requires that the equations be mathematically correct.

However, even a mathematically correct equation can be complete rubbish if it treats physical principles incorrectly.

Which brings us to the second definition of "proof", which is actually demonstrating it in the real world and providing raw data...

LeeroyJenkins

“I am very smart👌🏻, so much smarter that’s you🖐🏻, probably the smartest 👌🏻 person ever☝🏻.” - yuri


Why don’t you publish? You have a PhD. You know how the process works. I’m sure there is some masters or PhD student that you could work with to get published.

I've already asked him that. His stuff has probably been rejected by Wikipedia (they have strict rules against self-promotion) so it's unlikely any paper he writes will be accepted.

yuri_base

WSE are classic and eternal

They'll last as long as the people who think they're correct, and the server hosting on his website. And given his luck trying to convince people in the past 12 years (but refusing to provide proof), I wouldn't bet on it...

Quote

If unpowered flight is considered in a book, it's only dealing with parameters like AoA, Cl, Cd, surface area, etc. - exactly what the trick of WSE manages to avoid completely, since we don't know these parameters

But you know surface area. You can just lay out the suit on the ground and measure it!

You measured AoA with your vane, which is an extremely old device yet you treat it like it's a revolutionary invention.

Quote

Never in the history of aerodynamics, a problem of calculating a trajectory of a glider, standing on a tall cliff and stepping off of it, was needed

NASA never needed to calculate trajectories to land on fixed locations? Glider clubs also have accuracy competitions. You're not only claiming that this problem has never existed before, but additionally you're claiming that wingsuiters do not follow the same laws of aerodynamics as other gliders.

yuri_base

If such an aerospice engineer were under my supervision, I'd immediately fire them if they said something like the quote to me. God forbid they design some critical part in an airplane

Would you like my boss' email?

I have a simple question: what's the difference between a wingsuiter and the unpowered gliders that have been around for more than 100 years?

Also, please learn to spell

LeeroyJenkins

Why do you think 3.567 is the magic L/D ratio for wingsuit flight? What parameter of the wingsuit flight is that magic number for?

He's not going to answer that, it's too straightforward Also he's pretty much talking to himself now.

However, I can answer about the 1.41 L/D (which he thinks he discovered and is "magic") , it's trivially derived from 1/2 *rho*V^2*S*C_d if you treat C_d as a constant (which seems to be the limit of his abilities).

In this case, he's lucky because the induced drag component at higher speeds is small compared to profile drag. However, this value only assumes a straight-line glide and constant velocity. The path with the highest average horizontal velocity will be a cycloid similar to those derived by Jakob Bernoulli and Isaac Newton in the 17th century. (much harder to fly, though)

platypii

Based on your models, do you think we should be working more on increasing lift (surface area)? Or more work on decreasing drag? How do the wingsuit equations help inform which changes to try?

Unfortunately, the so-called "wingsuit equations" are wrong. However, doing both (increasing lift and decreasing drag) will help winguit performance.

Yuri claims that standard aerodynamic equations don't apply to unpowered gliding flight, but some of the best research has been done by NASA in the 1960s for lifting bodies which have glide ratios close to wingsuiters.

https://en.wikipedia.org/wiki/NASA_M2-F1

Unpowered aircraft have been around for even longer than powered aircraft. The NASA M2-F1 didn't even have any engines - if the only reason the drag polar is not applicable to wingsuiters is the presence of an engine, then NASA should have noticed very quickly. But they completed 77 test flights, and did 16 more flights with the M2-F2.

https://en.wikipedia.org/wiki/Northrop_M2-F2

It's even worse than that. Even if the wingsuit manufacturers didn't use scientific methods to design their suits (and I think they do), they've achieved much more than he has. His failure stands independent of whatever methods they use.

And he's just blaming most of the skydiving community for not accepting his genius. What I find laughable is his excuses for not providing proof.

SethInMI

He wants wingsuit designers to use a more scientific approach to design, including proper testing instrumentation.

I have no idea if wingsuit designers use a scientific approach to design. It would be better if they did, I agree. But Yuri's approach is not scientific and there are many flaws with his methods, not to mention the conclusions he reached.

SethInMI

and is left with just posting here.

throwing insults and calling wingsuit manufacturers names because they don't implement his (unproven) ideas is not very scientific...it's immature.

yuri_base

***Then just build one prototype and prove all the naysayers wrong by setting records. Can you?

Quote

The non-zero angle of incidence for both arm and leg wings? ("superterminal wingsuit", link in OP) I tried it ca. 2007-8 with the arm wings on my Phantom-1 and Vampire-2, it was promising, but I couldn't quantify it, because at the time I didn't have the instrumentation. I did it by sewing a second set of tabs going from armpits towards crotch, not along the sides (which is zero AoI).

Quote

Half-windgsuit with leg wing only? I tried it many times, starting with flying just the pants of my Prodigy back in 2005-6 and the advantages in speed were obvious. Also, eventually cut both P-1 and V-2 in half and flew the leg wings. Also, cut my V-4's arm wings completely after the zippers broke, and flew it like a Superman, with arm(s) stretched forward. Flew it in Eloy, Perris, San Diego, Lodi - no one is interested.

Those aren't records as far as I know. Put a flysight on and prove you can fly faster than anyone else.

Quote

I also won't ever participate in competitions since it's unthinkable to my dignity to be a gladiator - food for lions, puppet for the show. I don't have to prove anything to anyone.

I think it's because you'll probably lose.

Then just build one prototype and prove all the naysayers wrong by setting records. Can you?

yuri_base

And because they refuse to use even freely available ideas spoon-fed to them on the forums.

So ungrateful, aren't they? all they need to do is design, test, build, iterate, market, deliver and support the suits! You've already done all the work

yuri_base

(They've also raised the prices unilaterally to compensate for the new players on the market taking their slice of the pie.)

That's not usually how competition affects prices...

Quote

It's hopeless anyway, as I learned over the past 12 years.

If you're right, why don't you just build one for yourself and set records in speed, etc? If that happens I'm sure people will pay attention. Even better, start your own wingsuit company!

Quote

The solution of these is:

Kl = Vxs/Vs^3
Kd = Vys/Vs^3

With these adjusted coefficients, the wingsuit equations are

dVx/dt = g*V*(Kl*Vy - Kd*Vx)
dVy/dt = g*(1 - V*(Kl*Vx + Kd*Vy))

Now the unknown wingsuit parameters (wingloading mg/S and aerodynamic properties Cl, Cd) are "hidden" inside coefficients Kl and Kd, which can be easily calculated from sustained horizontal and vertical speeds.

Sigh. You treated C_l and C_d as constants. Which means if you use the equations that you used in the beginning:

D = 1/2 rho * V^2* S *C_d

Then yes, setting V to 0 will result in D = 0. Is this why you think the drag polar is invalid? Because of D = qSC_d?

Also, you said in the post:

Quote

We have 2 equations for 3 unknowns (k and Cl, Cd). But note that if we combine k with the Cl and Cd into "adjusted" lift and drag coefficients (no longer nondimensional)

Which is incorrect. k is known - it's 0.5*rho*S/m which is constant (rho = approximately 1.225 kg/m^3). So I'm not sure why you had to condense them to yet another set of coefficients.

It's understandable why D = 0 if V = 0 when you use that equation, but this drag term is not a constant - it's a condensed set of coefficients itself and is itself calculated with another equation. The drag polar was not derived from the equation you started with, which is why you got a different result. The equation you started with has been extremely simplified.

The V in this equation also denotes airspeed, and GPS measurements only measure ground speed.

Please refrain from personal attacks.

We use subscripts, e.g. C_L and C_d to denote lift and drag coefficients.

If you're a PhD then you should know how the peer review process works! (And if you're a PhD then you would also have published a few academic papers - so you're not really 'outside' the system are you?)

Quote

(And I strongly think, it never will.)

um...

Quote

Instrument noise is taken into account by L/D Magic's Kalman filters.

That's not how noise works...did you do any experiments during your Physics PhD?

Quote

In the whole Solar System.

Right... I'm done here. Good luck!

You can't have L isolated on one side of the equation then have it on the other side too. Same with D. Equations need to be equal (that's why they're called equations).

Quote

I feel no need.

Well, the burden of proof is really to prove that your "equations" are correct, as you haven't provided any proof that the prevailing academic thought on aerodynamics is wrong. And please stop using a dead person as proof. Anecdotes are not data.

The burden of proof should be on you as you're using them to instruct other wingsuiters and if they change their behaviour according to your possibly incorrect formulas, you should have the experimental data to back it up.

I don't have time to fully review your experimental methods, but without wind tunnel tests and only relying on wingsuit jumps with variable weather data (you did record all wind conditions, temperatures, pressures, and humidity, right?), with noisy instruments that have not been calibrated against laboratory references, and of course a very variable and very human test model, lack of convergence testing and instrument parameters, lack of error bars in your final graphs - there's a lot to work on.

Quote

L = W*(L/D)/sqrt(1 + (L/D)^2)
D = W/sqrt(1 + (L/D)^2)

Those are not equations. That's like saying 1=2.

Quote

I also invented the ways of measuring L/D and other aerodynamic parameters by using accelerometer and other sensors on a vane in undisturbed airflow.

I'd strongly suggest you publish your wingsuit equations in a reputable academic journal then, it would be good to see full methodology and get it through peer review.

I'd also recommend using Microsoft Equation or LaTeX when you typeset your equations to make the equations easier to read in your PDF.

Quote

Because wingsuits (and any nonpowered AC) do not have the goal (and means) to maintain the level flight

I never said it was. I said the goal was constant velocity, not constant altitude.

So if your force diagram does not balance to zero then your equations are already wrong. Because F = ma.

Again, I'm sorry for saying it like this. You seem fairly good at maths, but it's really quite unlikely that you've found a groundbreaking new equation. Wingsuiters, as well as any other form of gliding flight, are probably better off using standard aerodynamics equations. Blue skies!

Quote

No, even for sustained flight, lift is not equal weight, this is because the force diagram is different

All the forces need to balance to zero if it's moving at a constant speed (either straight and level, or a straight-line unpowered glide).

The only difference between powered and unpowered is that unpowered, you're using a bit of your weight vector (W*sin(theta)) to push against whatever drag you're experiencing. Your wing won't generate a force perpendicular to the earth, but the component of that force vector perpendicular to the ground (L*cos(theta)) needs to be equal to your weight, otherwise you will be accelerating rather than staying at a constant descent velocity.

I don't think you've explained very well WHY the standard drag polar is not applicable to unpowered flight.

Unpowered flight has been around for way, way longer than wingsuits have. Unpowered gliding flight has even been around for longer than powered flight.

I hate to break this to you but, your equations are wrong.

Quote

Unfortunately, the way they teach aerodynamics to aeronautical students, pounds this fallacy into their brain. From my observations over the years, aeronautical students and engineers do not have a solid grasp of fundamental physics concepts to recognize this fallacy. They fall into this pothole every. single. time.

I'm an aerospace engineer, and I don't know anyone who actually takes that plot as no more than a rough approximation of drag. The drag increase towards zero does not actually go to infinity, as it levels off at stall speed.

You say that drag is proportional to just the square of airspeed ( D~=v^2) but you've probably measured that drag with a constant angle of attack. The reason why the induced drag in the graph increases with decreasing speed is that you need to increase the angle of attack to maintain straight and level flight as you get slower. So yes, this graph is usually meant for powered aircraft in straight and level flight rather than gliding flight.

However, in gliding flight you still need to take this graph seriously. Why? Because your lift must still equal your weight! If you are flying your best glide angle, and your lift is not equal to your weight, then your speed will not be constant. You will be accelerating. As you yourself said, F = ma.

So, if lift is not equal to your weight, L - W = ma, which will accelerate you towards the ground. So if your definition of a good wingsuit flight is a parabolic trajectory towards the ground, I'm not sure how many people would agree with you.

On the other hand, I do agree with you that CFD in wingsuit situations needs to be taken with a large grain of salt. Low Reynolds numbers, large amounts of turbulence and large aeroelastic effects are not easy to simulate accurately even with the latest CFD algorithms.

One thing that might help is creating a Bode plot of the filters you're using - if you put them one after the other, you can calculate a transfer function, pass band, and the phase difference (lag).

As for the vertical speed, the derivative of a noisy signal will have the noise amplitude multiplied by the noise frequency, so the noise is greatly magnified and makes it quite a challenge

Fair enough. Why are you doing such heavy 3-stage filtering then? A simple low pass filter would suffice to eliminate barometer noise at the cost of bandwidth. In fact many of the MEMS barometers now have built-in low pass filters available. Do you know what's your end bandwidth after all the filtering?

A lot of the noise in the pressure readings in skydiving have nothing to do with the internal sensor noise, but rather from turbulence, motions of the skydiver, or things like opening the aircraft door. Since they do not originate from the sensor (more or less fixed time intervals) you'll want to filter them out anyway, and in the end you'll find out that no matter what barometer sampling rate you start with (well, at least beyond 10Hz), you'll end up with the more or less the same frequency response.

Both your replacements are now on the way. We're quite sorry to landersohn but he emailed us at 3:52am and 4:02am this morning, then an hour later started posting...by the time we woke up there was this post and a 1-star review...we need to sleep too if there are any issues please don't hesitate to contact us at [email protected]

20_kN

One of my largest complaints about modern alts

When I started skydiving I noticed that "modern" altimeters were nowhere near modern, which is the main reason why we started building our own www.aon2.co.uk) if you're interested in something more advanced