Derivation of the ETP formula
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Cracked it ! The equal time point is the the point where onward to the destination is the same time as return to the departure.
Using the symbols of the equation that means:
ETP/H = (D-ETP)/O (time = dist/speed)
so ETP/H = D/O - ETP/O (expanding RHS)
and ETP/H + ETP/O = D/O (adding ETP/O to both sides)
and (OETP + HETP)/OH = D/O (bringing the LHS to a common denominator)
so OETP + HETP = DOH/O = DH (multiplying by OH)
and ETP(O+H) = DH (factorising the LHS)
therefore ETP = DH/(O+H). QED.
Well it kept me happy for a couple of days !
Using the symbols of the equation that means:
ETP/H = (D-ETP)/O (time = dist/speed)
so ETP/H = D/O - ETP/O (expanding RHS)
and ETP/H + ETP/O = D/O (adding ETP/O to both sides)
and (OETP + HETP)/OH = D/O (bringing the LHS to a common denominator)
so OETP + HETP = DOH/O = DH (multiplying by OH)
and ETP(O+H) = DH (factorising the LHS)
therefore ETP = DH/(O+H). QED.
Well it kept me happy for a couple of days !
Last edited by puntosaurus; 27th January 2008 at 16:06.
Guest
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Good. Well done. Now do it for the engine failure case
(I did. Trying to be a total smart-Alec. And when I got to BGS was told that such questions never appeared for ETP/CTP!)
If you're looking for another couple of proofs to do how about Vmp=0.76Vmd or Vrange=1.32Vmd? Subject to jet/prop considerations of course....
Enjoy!
(I did. Trying to be a total smart-Alec. And when I got to BGS was told that such questions never appeared for ETP/CTP!)
If you're looking for another couple of proofs to do how about Vmp=0.76Vmd or Vrange=1.32Vmd? Subject to jet/prop considerations of course....
Enjoy!
Joined: May 1999
Aviation Qualifications: ATP+Mil
Posts: 27,398
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From: Quite near 'An aerodrome somewhere in England'
Ah, good old 1.32 (or, to be precise, 1.316074)!
When I went through CFS, the grounskool bod couldn't explain it and just said " 'cos it is!".
So I did a bit of basic mechanics and calculus:
Total drag = Zero Lift Drag + Lift Dependent Drag
i.e. D = AV**2 + BV**-2
To find the minimum drag speed, differentiate with respect to V and equate to zero:
dD/dV = 2AV – 2BV**-3 = 0
Hence VMD =(B/A)**-4
Power is the ‘rate of doing work’, i.e. Work/time or (Force x dist)/time
Hence Power = Force (in this case Drag) x velocity
P = DV
Thus P = AV**3 + BV**-1
Again, to find the minimum power speed, differentiate with respect to V and equate to zero:
dP/dV = 3AV**2 - BV**-2 = 0
Hence VMP = (B/3A)**-4
The relation between VMD and VMP is thusVMD = VMP x 3**-4
and therefore VMD = 1.32 VMP
Sorry if this isn't easy to read - this basic forum doesn't support subscript or superscript notation, hence I've had to use ** as 'to the power of', so **-4 means 'the fourth root of'.
A similar argument will give you the 0.76 answer - it's the reciprocal of the fourth root of 3, i.e. 0.7598357.
When I went through CFS, the grounskool bod couldn't explain it and just said " 'cos it is!".
So I did a bit of basic mechanics and calculus:
Total drag = Zero Lift Drag + Lift Dependent Drag
i.e. D = AV**2 + BV**-2
To find the minimum drag speed, differentiate with respect to V and equate to zero:
dD/dV = 2AV – 2BV**-3 = 0
Hence VMD =(B/A)**-4
Power is the ‘rate of doing work’, i.e. Work/time or (Force x dist)/time
Hence Power = Force (in this case Drag) x velocity
P = DV
Thus P = AV**3 + BV**-1
Again, to find the minimum power speed, differentiate with respect to V and equate to zero:
dP/dV = 3AV**2 - BV**-2 = 0
Hence VMP = (B/3A)**-4
The relation between VMD and VMP is thusVMD = VMP x 3**-4
and therefore VMD = 1.32 VMP
Sorry if this isn't easy to read - this basic forum doesn't support subscript or superscript notation, hence I've had to use ** as 'to the power of', so **-4 means 'the fourth root of'.
A similar argument will give you the 0.76 answer - it's the reciprocal of the fourth root of 3, i.e. 0.7598357.
Guest
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10/10 to BEagle! For a while I thpught only me and punto were the techno-anoraks
The proofs are irrelevant for pilots. Until as an instructor one's asked 'why?' Especially since the numbers are so odd-looking that you'd think they had to be type-specific.
The proofs are irrelevant for pilots. Until as an instructor one's asked 'why?' Especially since the numbers are so odd-looking that you'd think they had to be type-specific.
Joined: May 1999
Aviation Qualifications: ATP+Mil
Posts: 27,398
Likes: 857
From: Quite near 'An aerodrome somewhere in England'
If you're asked why, just ask the questioner whether he/she remembers O-level Additional Maths level calculus and O-level mechanics (statics and dynamics).....
Usually shuts them up!
Usually shuts them up!
Joined: May 1999
Aviation Qualifications: ATP+Mil
Posts: 27,398
Likes: 857
From: Quite near 'An aerodrome somewhere in England'
In level flight, assume L = W
At this weight, for minimum drag, it is necessary to fly at the maximum lift/drag ratio:
But L/D = CL / CD
The CL vs. α and CD vs. α curves for the aircraft are unique to that aircraft and independent of weight.
This means that, to fly at minimum drag speed, the speed must be that which maintains level flight for the weight of the aircraft at the time at the (CL / CD) max value of Angle of Attack. Thus if the weight changes, so will minimum drag speed.
For jet aircraft, best endurance speed is at (CL / CD)max, whereas best range speed is at (CL**-2 / CD)max. Best turn rate is at CL max . Hence in the F4 we flew by AoA for each requirement - no worrying about the exact speed, just decelerate to fly level at the appropriate AoA for either range or endurance. For best turn, full A/B and pull to the best turn AoA at the speed corresponding to max permitted +G!
At this weight, for minimum drag, it is necessary to fly at the maximum lift/drag ratio:
But L/D = CL / CD
The CL vs. α and CD vs. α curves for the aircraft are unique to that aircraft and independent of weight.
This means that, to fly at minimum drag speed, the speed must be that which maintains level flight for the weight of the aircraft at the time at the (CL / CD) max value of Angle of Attack. Thus if the weight changes, so will minimum drag speed.
For jet aircraft, best endurance speed is at (CL / CD)max, whereas best range speed is at (CL**-2 / CD)max. Best turn rate is at CL max . Hence in the F4 we flew by AoA for each requirement - no worrying about the exact speed, just decelerate to fly level at the appropriate AoA for either range or endurance. For best turn, full A/B and pull to the best turn AoA at the speed corresponding to max permitted +G!
Last edited by BEagle; 30th January 2008 at 15:53.
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llanfairpg
It can be proved mathematically. I think. And I know I'm going to regret this.
Cd=Cd0+kCl^2
Divide through by Cl to give
(Cd/Cl) = Cd0/Cl + kCl
Assume, for illustrative purposes, that Cl = a . alpha (ie symmetric, away from stall - variable a is the slope of Cl vs alpha curve)
(Cd/Cl) = Cd0/(a.alpha) + k.a.alpha
(Cd/Cl) is a minimum with respect to alpha (ie Cl/Cd is maximum) when you differentiate (Cd/Cl) w.r.t alpha and set the result to zero.
Do this and you get k(a.alpha)^2 = Cd0 or kCl^2 = Cd0 at best Cl/Cd.
But lo and behold....kCl^2 = Cd0 is exactly the condition at minimum drag. (This is intuitive but can also be proved mathematically). QED
Where's Genghis when you need him....
It can be proved mathematically. I think. And I know I'm going to regret this.
Cd=Cd0+kCl^2
Divide through by Cl to give
(Cd/Cl) = Cd0/Cl + kCl
Assume, for illustrative purposes, that Cl = a . alpha (ie symmetric, away from stall - variable a is the slope of Cl vs alpha curve)
(Cd/Cl) = Cd0/(a.alpha) + k.a.alpha
(Cd/Cl) is a minimum with respect to alpha (ie Cl/Cd is maximum) when you differentiate (Cd/Cl) w.r.t alpha and set the result to zero.
Do this and you get k(a.alpha)^2 = Cd0 or kCl^2 = Cd0 at best Cl/Cd.
But lo and behold....kCl^2 = Cd0 is exactly the condition at minimum drag. (This is intuitive but can also be proved mathematically). QED
Where's Genghis when you need him....
