Aces High Bulletin Board
General Forums => Aircraft and Vehicles => Topic started by: funkedup on July 31, 2003, 02:05:32 PM
-
I notice there seems to be much discussion and confusion about this.
For an airplane in a vertical dive,
Acceleration = g + (Thrust-Drag)/Mass
In general, the "g" term is significantly larger than the "(Thrust-Drag)/Mass" term. Unless the aircraft are drastically different (e.g. WWI biplane vs. X-15), you aren't going to see huge differences in dive acceleration.
At relatively low speeds (below the level maximum speed), thrust exceeds drag, so more mass is a bad thing.
At relativiely high speeds (above the level maximum speed), drag exceeds thrust, so more mass is a good thing.
-
Originally posted by funkedup
I notice there seems to be much discussion and confusion about this.
For an airplane in a vertical dive,
Acceleration = g + (Thrust-Drag)/Mass
My "I don't believe you" light just lit.
If you do the unit calculation...
Acceleration = distance/time^2
Drag = ??
Mass = mass :)
distance / time^2 = distance / time^2 + ( ??/mass).
You end up with some really screwy units there...whatever it is isn't acceleration.
-
Puck start with F=MA.
G = 32.2 FPS^2
F = Weight + T - D
Weight = (mass * G)
A = (M * G + T - D) / M
Or as funked said A = G + (T - D)/M
Btw funked had never view it in quite that equation form b4, does make it intutiivly obivous that accelerating in a dive at max level speed, is like droping a ball in a vacum.
And in our planes T is rairly much more than 1/4 weight at climb speeds.
HiTech
-
Something I cooked up with Paint (yes, I am bored) that might be a bit easier to understand:
(http://home.earthlink.net/~mathmanahs/diveaccel.jpg)
As you can see, everything ends up in the same units at the end :)
-
Much better. Thanks.
That'll learn me to do higher math at work, eh?
-
Accel is (thrust(lbs)- drag(lbs))/slugs
slugs = weight(lbs)/32.2
You can do it in metric, but in that way lies madness.
Dive accel is just a variation on the above. Part of the aircraft's weight(lbs) is acting as thrust(lbs). Sine or Cosine (never can remember which is which) of the dive angle times the weight. So, the steeper the dive the more of the planes weight is acting as thrust.
You also have to take speed into account. Engine thrust(lbs) = P% * 375/TAS * hp
Prop efficiency is starting to drop at high TAS as well, so things get even worse the higher the dive speed.
So at high dive speeds engine thrust is dropping off, and thrust due to weight is becoming a more significant part of the equation.
The steeper and longer the dive, the more the advantage goes to the heavier fighter. Explains why the Bf 109 or Fw 190 could initially out accellerate the USAAF fighters, but the americans could run them down in an extended dive.
Greg Shaw
-
It would appear from that formula that the airplane with the better level acceleration will have the advantage up to it's max level speed. Beyond that point, the airplane with the most weight and least drag will pull away ... albeit dive acceleration beyond the airplanes max level speed will always be slower then 32.3 fps ^2. That explains why it's tough to pull away from a Spit IX or N1K2 unless it's a really long dive.
-
Sable that's right.
The max level speed will be a little lower than the vertical flight (thrust = drag) speed, due to induced drag.
(1g normal for level flight vs 0g normal in vertical dive)
But it will be very close, because induced drag is a pretty small percentage of drag at speeds that high.
-
would the aircraft with a better thrust to weight ratio be the aircraft that is faster accelerating in a dive? as in the better the ratio the faster that aircraft achieves it max speed or is it the other way around? (drag being equal for arguements sake)
-
Hazed look at the equation. The acceleration in a vertical dive depends on Thrust/Mass and Drag/Mass. They both have equal influence.
-
Ok now I tried the solve this and all I got was this.
A=Ag+(At-Ad)
Ad(12)=583
:confused:
-
P-38J
16500 lbs, 3200 hp @ 25,000 ft, 300 mph TAS, P% 80
thrust = 3200 * .8 * 375/300 = 3200 lbs
drag = 1275 lbs (I'm too lazy to go over this formula right now, I have a spreadsheet that calculates all this on the fly)
slugs = 16500 / 32.2 = 512
(3200 - 1275) / 512 = 3.75 fps accel in level flight @ 25,000 ft
45 deg dive
thrust = thrust + (.707 * 16500) = 14865 lbs
(14865 - 1275) / 512 = 26.5 fps accel
Not quite that simple, you can't go from level flight to 45 deg dive instantly. You also can't dive at full power, you need to throttle back and override the prop, or risk overspeeding and self destructing the engine. 22-23 fps is probably about the limit.
Greg Shaw
-
A diving plane produces less lift and therefore less drag than in level flight. So in a dive the speed where thrust=drag should be faster than in level flight, correct?
-
I think your basically saying top speed in a dive is faster than is attainable in level flight? If so then I agree, yes. Diving is faster.
-
Originally posted by Regurge
A diving plane produces less lift and therefore less drag than in level flight. So in a dive the speed where thrust=drag should be faster than in level flight, correct?
A little bit faster.
-
Hi Regurge,
>A diving plane produces less lift and therefore less drag than in level flight. So in a dive the speed where thrust=drag should be faster than in level flight, correct?
If only the wing would count, yes.
However, the fuselage is usually aligned with the air stream at top speed at full throttle height. So at zero lift, you get less drag from the wing but more from the fuselage
Depending on the exact conditions, minimum drag for a given speed might be achieved anywhere between 0 G and 1 G.
Regards,
Henning (HoHun)
-
Doh I just repeated what funked said replying to sable. I must have missed it.
I see what you're saying hohun. So mimimum total drag would have positive wing AoA and negative fuselage AoA (assuming positive wing incidence). I suppose thats all pretty insignificant compared to the parasitic drag at high speed though.
-
Hi Regurge,
>I suppose thats all pretty insignificant compared to the parasitic drag at high speed though.
I believe you're right. The effect seems to be much more pronounced during "unloading" - pushing to 0.5 G may be more energy-efficient than unloading to 0 G as a result. If you get fast, parasitic drag takes over and unloading isn't worthwhile anymore.
Regards,
Henning (HoHun)