Author Topic: Climb Vrs Acceleration Part Deux AFDU data (Read 702 times)

F4UDOA · « **on:** July 18, 2002, 05:20:00 PM »

Someone care to explain this chart?

How is it possible that the Tempest which climbs like a rocket only accelerates like a ThunderBolt??

Whats up with this??

Ain't I a pest;)

Holden McGroin · « **Reply #1 on:** July 18, 2002, 06:28:59 PM »

Rate of Climb = 60 x V x Angle of Climb

Angle of Climb = [550 X Prop Eff x sq rt of (Airdensity / 2)]

all divided by {Gross WT/ Power} x sq rt of {GWT/ Wing Area}

inhale

then subract total aircraft drag coeff - 1 / { pi x AR }

AR = wing aspect ratio....

so by equating climb to hp, you only neglect aspect ratio, prop effeciency, air density, gross weight, wing area, and overall drag.

otherwise, it is a perfect ratio.

Accelleration related only to the balance / imbalance of thrust vs drag.

Sclew · « **Reply #2 on:** July 19, 2002, 02:00:30 AM »

Prop optimization.

You can build a prop to climb, or to go fast. If you build it to climb you get good low speed effeciency and initial acceleration. If you build it to go fast you get shoddy low speed effeciency but better thrust at high speeds and more efficiency at higher alts.

Add more HP, and as HT said before, the issue only gets much worse.

Personally I am wondering what tests the used for "initial acceleration". The Moostang should not be that far above the Tempest V in any test I can think of.

Naudet · « **Reply #3 on:** July 19, 2002, 03:40:34 AM »

DOA, do u have a better quality scan? It's pretty hard to read aircraft designations, especially in turning circle diagram.

Edit: btw were is that page exactly from, and were can i get it?

hitech · « **Reply #4 on:** July 19, 2002, 10:59:12 AM »

DOA Re Read the climb section, the tempest was the worst except for the Thunder bolt, appears acceleration is the same order.

F4UDOA · « **Reply #5 on:** July 19, 2002, 01:59:03 PM »

Naudet,

It was sent to me from a gent in the UK who also sent me the F4U-1 and F6F-3 trials at Boscombe downs:D. I will post the whole thing so you can have the entire report.

Hitech,

The climb was definitly not in that order. I didn't post that chart yet.

Here are the other pages.

funkedup · « **Reply #6 on:** July 19, 2002, 02:31:48 PM »

Here's how the whole climb/accel thing works.

* = multiplicaton
d[]/dt = differentiation operator, e.g. df/dt = first derivative of f with respect to time.
t = time
E(t) = energy as a function of time.
P(t) = excess power as a function of time.
v(t) = TAS as a function of time.
h(t) = height as a function of time.
m = mass
g = gravitational constant

Total energy of the airplane, assuming the mass of the aircraft is not changing.
E(t) = 0.5 * m * v(t) * v(t) + m * g * h(t)

P(t) is defined as dE(t)/dt. The definition of power is the rate of change of energy with respect to time.

Therefore:

P(t) = d( 0.5 * m * v(t) * v(t) + m * g * h(t) )/dt
P(t) = m * v(t) * dv(t)/dt + m * g * dh(t)/dt

dv(t)/dt is defined as acceleration.
dh(t)/dt is defined as rate of climb.

Assuming level flight, dh(t)/dt = 0
P(t) = m * v(t) * dv(t)/dt + m * g * 0
P(t) = m * v(t) * dv(t)/dt
dv(t)/dt = P(t) / (v(t) * m)
That is, at any instant in time, acceleration in level flight is equal to excess power divided by the product of mass and TAS.

Assuming constant speed flight, dv(t)/dt = 0
P(t) = m * v(t) * 0 + m * g * dh(t)/dt
dh(t)/dt = P(t)/(m*g)
That is, at any instant in time, the sustained rate of climb is equal to excess power divided by the product of mass and the gravitational constant.

The only things I used to get this are the definition of energy, the rules of calculus, and the assumption that g and m are constant. Even if you assume g and m are not constant and crunch the numbers, the contribution of those variations is vanishingly small for the type of aircraft we are talking about.

F4UDOA · « **Reply #7 on:** July 19, 2002, 05:20:50 PM »

Gents,

First.

I get the point that physics says it so.

However if you look at the climb charts I've posted compared to the accleration charts they don't match. Point being that I believe there are some variables that the current physics model does not take into account.

Case and point is the Tempest in the AFDU. It climbs almost 5K per minute but does not acclerate well compared to the others in this chart.

Why? I dont think the match is wrong. But somethings wrong.

Badboy · « **Reply #8 on:** July 19, 2002, 07:01:15 PM »

Quote

Originally posted by F4UDOA
However if you look at the climb charts I've posted compared to the accleration charts they don't match. Point being that I believe there are some variables that the current physics model does not take into account.

Nope, the theory is good, but those charts are only intended to compare the different aircraft with respect to one single performance attribute at a time! The climb and acceleration data cannot be cross referenced, they were never intended to be used that way. The climb data for the tempest was for a speed in excess of 180mph while the acceleration data is “initial acceleration” which may mean that the acceleration was averaged over a time interval as the aircraft accelerated from rest.

Hitech has already explained this to you in his very first response to your previous thread…Here it is, just in case you missed it:

Quote

Originally posted by hitech
F4UDOA, Your taking the word instantious wrong, meens at a single speed for acceleration. I.E. The acceleration from 169 to 170 is different than acceleration from 200 - 201.

Accelerations would be in a unit like Mph change in 1 sec.

Also climb rate at 200 is different than climb rate at 169.

The chart your looking at gives a climb rate at 1 speed, and an average acceleration over different speeds.

When we say Climb rate is always perportional to acceleration we meen that if you have an accelerations of 10 mph per sec and a climb rate of 5000 fps, at 160 mph.

Then if acceleration at 200 mph = 5 mph per sec the climb rate at 200 mph must be 2500.

Note you must hold a constant 200 during climb by either raising or lowering the nose.

And they will always be directly perportional.

HiTech

That’s exactly what’s happening in your second example, it isn’t because anything is wrong with the charts or the physics, you are simply trying to use the charts in a way they can’t be used!

Quote

Originally posted by F4UDOA
Case and point is the Tempest in the AFDU. It climbs almost 5K per minute but does not acclerate well compared to the others in this chart. Why? I dont think the match is wrong. But somethings wrong

That’s because the climb rate for the Tempest in those charts is at the highest speed of any of the other prop fighters, if you were to compare acceleration starting from that speed, it would have exactly the same ranking as it does for climb. The aircraft that accelerate much better at the lower speeds would compare less favorably to the tempest at the higher speeds. If the charts were meant to be cross-referenced they would reflect that, but they weren’t. So, it comes back to the same thing, the charts you have posted simply can’t be used the way you are trying to use them, that's the only reason they don't correlate the way you expect them to... Trust me, nothing is wrong!

Badboy

bozon · « **Reply #9 on:** July 19, 2002, 07:57:06 PM »

funkedup, your definition of P(t) already includes the drag hidden in it. using this definition doesn't show you the dependancy on speed and the aerodynamics, which are the true interesting parts.

the "excess power" as you described it depends on drag just as much as on engine output and prop efficiency. just looking on engine power to weight ratio won't tell you much.

Bozon

gripen · « **Reply #10 on:** July 20, 2002, 02:41:09 AM »

I've seen these charts before and IIRC these were published in the Flight or Aeroplane magazine.I don't know if the charts are very accurate. I have test data on Mustang which claims that it reached it's best climb speed around 195mph IAS at WEP.

gripen

funkedup · « **Reply #11 on:** July 20, 2002, 07:12:12 PM »

Quote

Originally posted by bozon
funkedup, your definition of P(t) already includes the drag hidden in it. using this definition doesn't show you the dependancy on speed and the aerodynamics, which are the true interesting parts.

the "excess power" as you described it depends on drag just as much as on engine output and prop efficiency. just looking on engine power to weight ratio won't tell you much.

Bozon

No toejam Sherlock.

bozon · « **Reply #12 on:** July 20, 2002, 07:52:51 PM »

Quote

No toejam Sherlock.

it's elementry my dear Watson.

Bozon

niklas · « **Reply #13 on:** July 21, 2002, 06:19:28 AM »

I think you need first the definition of inital acceleration.
Is the the acceleration when beginning a dive, pushing forward the stick? Or level flight, starting from cruising speed? or 0.7 v_max??

Very interesting to see that the climbrate of the AH TempestV matches the +12lb Temepst II with Centaurus engine (2600-3000PS??) btw...

niklas

Holden McGroin · « **Reply #14 on:** July 21, 2002, 03:00:54 PM »

Initial accelleration is the accelleration that happens initially.

If we are talking level acceleration, and that is the only acceleartion that horsepower has anything to do with, then just after you max out the throttle, initially, that is initial acceleration.

Otherwise, a rock initially accelerates at 32.2 ft/sec sq

The thing that makes the difference is aerodynamics. Drag at whatever the climb speed is varies greatly among all these designs. The aerodynamic qualities of the wing, prop, and fuselage make a great deal of difference. Excess horsepower is not the only climb factor; weight and aerodynamics balance the equation.