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Vy equals 1.32*Vx

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Vy equals 1.32*Vx

Old 17th Jan 2018, 14:39
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Wink Vy equals 1.32*Vx

Lets first agree on the following terminology:
Vy = speed for max rate of climb = speed for max range
Vx = speed for max angle of climb = max lift over drag = min drag
Furthermore it's possible for a jet engine, after a few assumptions and approximations, to mathematically conclude from the drag formulas that Vy=1.32*Vx.
All this seems to be generally accepted and I also find it relatively true when looking at speeds for the B737 and the A320 at lower altitudes. I have not been able to find the relevant data for high altitude.
The question is what happens at high altitude and why? The Oxford ATPL text claims Vy will decrease and eventually equal Vx. Is that true or is the factor 1.32 relevant for all altitudes?
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Old 17th Jan 2018, 16:38
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The higher you climb, performance decreases until it necessarily reaches a point when you you cannot climb any more, i.e., rate and angle are both zero and the best you can do is level flight.

If you get even a knot off Vy, rate becomes negative.

If you get even a knot off Vx, angle becomes negative.

You see why, in this situation, Vx has to equal Vy?
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Old 17th Jan 2018, 17:14
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Given enough thrust, a jet-powered aircraft could reach ‘coffin corner’, an altitude where controlled level flight is only possible at one speed. This altitude is the ‘absolute ceiling’.

At an altitude just below the absolute ceiling, flight at different speeds is possible but the available range of speeds will be small. A typical definition is that the ‘service ceiling’ is that where a rate of climb of 100fpm is still possible. It follows, therefore, that Vy and Vx will converge as altitude increases, eventually meeting at the absolute ceiling.

Sorry Vessbot, I think I crossed with your post.
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Old 17th Jan 2018, 18:12
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Ok, this all sounds great but what about the claim that Vy = 1.32 * Vx when eventually Vy = 1.00 * Vx? I just assume the factor is not decreasing linearly with altitude to absolute ceiling or is it? What's the physics behind this?
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Old 17th Jan 2018, 18:31
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Vx = speed for max angle of climb = max lift over drag = min drag
Max L/D ratio is only the same as VMD in level flight.
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Old 17th Jan 2018, 19:21
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eckhard

What you say is true, but I think it's belaboring the point. If the OP is asking about Vx/Vy type stuff, then I think they're still trying to figure out how the performance-limited ceiling works; and talking about hitting a Mach ceiling instead, only introduces more confusion at this point. It's a more basic question.

In figuring out why Vx and Vy converge, I'd advise ignoring Mach and imagining a lower performance airplane that will climb until excess thrust and power reach zero.
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Old 17th Jan 2018, 20:45
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I agree that introducing Mach number would be confusing, which is why I didn’t mention it!
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Old 17th Jan 2018, 22:39
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Yes you did, apparently without realizing it. "Coffin corner" means the speed-altitude point where the increasing stall speed converges with the decreasing Mach limit. There, you can only fly at one speed without exceeding one or the other of those limits. But it has nothing to do with running out of excess thrust or power, i.e. the performance ceiling.
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Old 17th Jan 2018, 22:49
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SO the only way out of that situation is to descend? Just checking...
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Old 17th Jan 2018, 22:55
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OK Vessbot, a fair point and well made!
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Old 17th Jan 2018, 23:20
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Originally Posted by joohaan View Post
Ok, this all sounds great but what about the claim that Vy = 1.32 * Vx when eventually Vy = 1.00 * Vx? I just assume the factor is not decreasing linearly with altitude to absolute ceiling or is it? What's the physics behind this?
I missed this post earlier. Yes, 1.32 is only true at sea level, and only true as an approximation. There is no exact mathematical law by which it reduces to 1.00
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Old 17th Jan 2018, 23:25
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Originally Posted by Pearly White View Post
SO the only way out of that situation is to descend? Just checking...
Yes. Interesting point about the U-2, which lived in coffin corner: separately from the wing aerodynamics, at its altitudes, idle power on the engine climbed so high that it met with the turbine temp limit of max power, so there was only one power setting. (One might call it a separate "coffin corner" of the engine.) Anyway, because of that, it could not reduce power and the only way to descend was to add drag!
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Old 17th Jan 2018, 23:41
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Originally Posted by joohaan View Post
Lets first agree on the following terminology:
Vy = speed for max rate of climb = speed for max range
This is true, and it is also the minimum total drag speed (maximum excess *power*)

Vx = speed for max angle of climb = max lift over drag = min drag
The bolded part is NOT true. Speed for best climb angle is NOT the minimum drag speed - it's the speed with the greatest excess THRUST. For a jet this may be close to (but still slightly less than) the minimum drag speed because jets have almost constant thrust at all airspeeds. But for anything with a prop the thrust reduces with airspeed, so the best climb angle speed will be significantly LESS than the minimum drag speed.
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Old 18th Jan 2018, 19:33
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For a jet this may be close to (but still slightly less than) the minimum drag speed because jets have almost constant thrust at all airspeeds.
This is only true for "pure jets" - and even then only to a first approximation. Fan jets lapse (lose) net thrust with forward speed, the larger the bypass ratio, the greater the thrust lapse with forward speed. The efficiency improvement of higher bypass ratios is large enough to overwhelm the effect of this thrust lapse, so overall cruise efficiency still improves, but it is a major reason why cruise speeds for jet airliners has decreased slightly compared to 50 years ago (and why fighter jets - especially those with supersonic capabilities - use very low bypass engines).
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Old 18th Jan 2018, 20:04
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Absolutely. I was trying to keep it simple!
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Old 19th Jan 2018, 06:05
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The question why remains unanswered. What are the assumptions made making the eqation true only att sea level? Maybe it is true as long as there is enough thrust to reach the speed 1.32*Vx?
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Old 19th Jan 2018, 14:02
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I’m no engineer, but I think the original equation is not valid at all.

As others said, Vx is the speed of max excess thrust and Vy is the speed of max excess power. They do not depend only on aerodynamics, but also on engine/prop characteristics.

The famous 1.32 factor is the relation between “min drag” speed and “best range” speed for a jet
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Old 24th Jan 2018, 14:41
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Yes, the 1.32 factor is the mathematical ratio between Vmd and LRC expressed as IAS values. (Because drag is proportional to IAS) It is also therefore valid at any altitude. In real life there is a few knots variation but thats up to an aeronautical engineer or test pilot to explain why! Vx & Vy maybe coincidentally similar but are different things.
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Old 25th Jan 2018, 08:24
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The factor 1.32 applies if the thrust is constant and the drag polar is 'parabolic'. A parabolic drag polar can be written as:
CD = a + b*CL^2
where a and b are constants, CD=drag coefficient and CL=lift coefficient

Constant a defines the parasitic drag which is proportional to airspeed squared.
Constant b defines the lift-dependent (or 'induced') drag which is inversely proportional to airspeed squared.

Most airplanes exhibit a nearly parabolic drag polar at low speed and low altitudes where Mach effects are insignificant. At high subsonic Mach numbers the parasitic drag coefficient is no longer constant but increases rapidly with increasing Mach number, so the drag polar is no longer 'parabolic' and the 1.32 factor no longer applies.

Last edited by Gysbreght; 25th Jan 2018 at 08:38. Reason: "coefficient" added in last sentence (in italics)
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Old 25th Jan 2018, 10:52
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Does anyone know the maths behind the constant 1.32 ?
Is it the result of differentiating the previously mentioned formula. The nearest guess that I have made is that it is the 4th root of 3.
.
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