you can NOT take the V out and write:
(V/2G)d(V)/dt
Otherwise you’re right in saying that there’s an acceleration as V^2 is increasing and d(V^2)/dt is positive.
I can’t? Math 101;
Its a matter of simplification has nothing to do physics behind it. v*dv/dt = 0 if a = 0
345m/s is not M0.9 at any altitude unless it’s a very hot day at sea level.
MiG; engineers say othervise. MiG-29;
[ATTACH=CONFIG]229939[/ATTACH]
Also General dynamics engineers;
F-16@22000 lbs has 1000 feet per second = 304,8m/s @ M0,72 = 247m/s. Note that I circled STD DAY conditions.
[ATTACH=CONFIG]229940[/ATTACH]
Or McDonnell Douglas engineers of F-18;
[ATTACH=CONFIG]229941[/ATTACH]
Not to mention Boeing engineers of F-15, Sukhoi engineers of Su-27? But they are ALL wrong and you the lukos the infallible is right, right?
The equation you’ve written actually gives the rate of change of V^2 not acceleration. There are simpler ways of estimating acceleration. I.e. the specific excess thrust, which is actually:
They are indifferent from energy gain AND MATHS (see above) perspecive, I don’t know how one can be so thickheaded to comply with this.
So once you use that excess power to achieve maximum climb rate, V is no longer equal to 309m/s, otherwise a climb rate of 345m/s would be impossible.
Still you are not getting the point AT INSTANT V=309m/s, MiG-29 achieves 345m/s climb rate. There is no “no longer” thingie about it. Its irrelevant if this is physically possible or not; this is not “MiG-29 can climb 345 m/s @V=309m/s” statement. It is “MiG-29 has excess power equivalent to potential energy gain of 345m/s climb rate”.
That is why you are completely wrong in your following calculations:
Vertically,in a stable climb, (T-D)sin(Aoc) = ma x sin(AoC) = W = mg
ma x sin (AoC) = mg
a x sin(AoC) = g
Even going completely vertical at 90 degrees; an object with airpseed of 309m/s cannot have greater than 309m/s climb rate; but MiG-29 is known to have 345. So Your climb angle calculations are invalid due to (T-D)sin(AoC) = W cannot be valid;
Since you like accurate numbers, let me explain this way;
(T-D)/W*V = 345 when V = 309m/s;
(T-D) = W*345/309;
(T-D) = 1.116*W
You have to take arcsin of 1.116 which you cant, and due to you ignore the validity for input/output range, you take the the arcsin of 1.116^-1 (0,896) which gives out your 63,64 degrees, which is WRONG.
It’s not really that valid. If you look at the equations above, it’s using the wrong ratios and therefore getting the wrong answers.
SO YOU CLAIM TWO EQUATIONS PLUS THREE CLIMB RATE GRAPHS ON MIG-29 FLIGHT MANUAL AERODYNAMICS BOOKLET IS WRONG????
Andraxxus is wrong, Amiga500 is wrong, 800+ engineers in MiG design bureu are also wrong, they designed MiG just by pure luck, only lukos is right. Why? because he is right and he also has PhD.
Perhaps its you who is using wrong ratios and getting wrong answers?
You need to show your working from first concepts because at the moment it just look plain wrong.
Like I said, I don’t go vectoral calculations; I go from energy calculations, take L=W and dE/dt = (T-D)*V. SEP is irrelevant of W; dividing both sides will give dH/dt = (T-D)/W*V as SEP or inst. climb rate;
Then I ratio them, assuming same V; so dH1/dH2=(T1-D1)/(T2-D2)*(W2/W1).
W can be ratioed accurately; but I use, (T1/T2)-(D1/D2) in place of (T1-D1)/(T2-D2); I know its inaccurate; but better than nothing.
This is where I disagree, V will not be the same and cancel and must be recalculated based on some kind of force balancing or account of a maximum on a SEP graph. Assuming the same Cd, or same climb angle (AoC) is fine but V will change and will likely be greater for aircraft with a larger T-D value. There are also problems with the mathematics of how you’re doing the ratio besides that and because of it. Because the different optimal V will also lead to a different optimal D, so it won’t move linearly with A. .
You are right if you are absolutely through and exact; However from experience; most aircraft achieve this around same airspeed (MiG-23@M0,85 MiG-29@M0,9 F-16@M0,82 Su-27@M0,84 etc etc). Its a valid assumption to simply ratio things.
The bit in bold. Assuming V is the axial velocity, you obviously can’t get climb rates >300m/s if V is limited to M 0.85 @SL and you aren’t even going directly up.
In fact, its pretty logical if you look at from energy gain point of view; A MiG-29 @12800 kg has 345m/s climb rate at M0,9 as its manual suggests; Even if climb angle is 90 degrees; Vsin(theta) will prevent actual rate of climb to be greater than 309m/s. So what happens? 309 of the 345m/s excess power is wasted on vertical climb, and remaining 36m/s is used for acceleration while conducting vertical climb.
To make it more scientificc; typically this is where this formula has its use;
Divide both sides by W; assume W doesn’t change with respect to time so move it out of differantial part; eqn becomes
(T-D)*V/W = dH/dt +d/dt(V^2/(2g)
You know MiG-29’s SEP (T-D)/W*V = 345 m/s and max current rate of climb is 309m/s as limited by Vsin90;
345m/s – 309m/s + (1/2G) dV/dt*V where dV/dt is acceleration, positive indicating MiG-29 is accelerating in a 90 degree climb; math will follow to find the exact value.
SEP gives impressive statistics about aircraft performance; take dH/dt = 0, and it will give you level flight acceleration of the aircraft by;
345m/s *2G/V = acceleration@M0,9 for example. Check the formula by its two dimensions m and s; (m/s)*(m/s2)/(m/s) = m/s2 and see it to be true.
Assuming you’re just using that as a starting point for the climb, calculating the increase in SEP at this speed from ratios is incorrect because if I take a simple situation where:
T = 10
D = 5
W = 5(T-D)/W = 1
If I increase T by 50% and D and W by 10%
(T-D)/W = 1.72 = [(15-5.5)/5.5]
1.72 is nowhere close to (1.5-0.1)/1.1 = 1.27 out by >26%.
W can be ratioed; but you are most definitely right that T and D can’t be ratioed that way due to addition/subtraction, we cannot get it out of paranthesis and group as a ratio. However as I don’t know T = 10 and D = 5; I have no means of implementing this; This is the prime reasion why I rated accuracy so low, because margin of error is huge(not because I assumed Rafale has same Cd as M2k); from numeric analysis POV, this should work as long as T is close to D; and margin of error will increase with their differences increase.
I assumed 60 degrees as best angle of climb (BAOC) because that seems to be roughly what fighters seem to use. You’d do well to go through that calculation.
Its still wrong, however because you are still taking this as actual climb; see MiG-29 example I mentioned and calculated above; source
[ATTACH=CONFIG]229927[/ATTACH]
I added some number and lines in paint don’t bother them; It was for another topic, too lazy to reupload the clean image.
Explain these figures with your methodology and I will go with your calculation;
M0,5; V= 171m/s -> 180m/s climb rate.
M0,7; V= 240m/s -> 280m/s climb rate.
M0,9; V= 309m/s -> 345m/s climb rate.
Assuming you accept MiG engineers are capable enough to provide correct climb rates for the aircraft they designed; tell me; what BAOC would MiG-29 require? 60degrees? 90degrees? 90+? Or perhaps its time you accept your methodology is wrong when talking about “climb rate” that means “specific excess power”.
PS; You made me notice an interesting thing; MiG-29s Vy*is always greater than the V*sin90 until M0,9, meaning it could go vertical climb and still accelerate at below M0,9 airspeed @ S/L, 12800kgs.
I’m referring to a zoom climb not a sustainable climb.
Indeed you are. and problem is, even in zoom climb, your energy gain will wont change. In my calculation you will gain X amount of PE and 0 amount of KE. If you go through long version, You will gain X+Y amount of PE and lose Y amount of KE; net result will still be X.
Well the word is confusing because they’re actually talking about something that’s sustainable at that point in time.
Yes! Finally we are talking in the same language. Its called instantenious because the instant aircraft gains 1 meter altitude, there will be change in air density, so all thrust drag and weight will be affected; as aircraft gains altitude, it will lose climb rate even though airspeed stays the same.
The problem is that whilst max climb rate is given by max excess power, given by:
[(F-D) x V]/W
The V in the equation is not the climb rate. Even in the equation pertaining to a Mirage 2000, the V is not the climb rate, the result of the whole equation is. That V which delivers maximum climb rate will change from aircraft to aircraft in a none linear manner. SO plugging in 285m/s V is complete crap and deserves, Accuracy 0/10.
I NEVER said that. V is the velocity; Assuming same V (which pretty valid for max climb rate, always around M0,85 and S/L); if you ratio New vs Old at same speed;
NewClimb/OldClimb = New(F-D)/Old(F-D)*(OldW/NewW) and “V” will cancel out.
What I assumed is Cd will be same between New and old, so D relation will be dependent on wing area alone.
As I have no initial data; I used “(NewF/OldF)-(NewD/OldD)” instead of “New(F-D)/Old(F-D)” Inaccurate I am fully aware of it, (hence why i called it so inaccurate; it would have much greater accuracy othervise).
Now explain how that applies. It’s just stating this with different symbology:
same as this actually; but as dV/dt = 0 you have to ignore KE change. If you dont; aircraft will trade PE for KE, and you will go through long and useless calculations to find same dE/dt or dH/dt, more likely you will make a mistake in your assumptions or calculations.
Sign In
