Author Topic: How to regain lost E? (F4U)  (Read 5382 times)

Offline dtango

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Re: How to regain lost E? (F4U)
« Reply #60 on: August 06, 2011, 03:56:44 PM »
What excess power? HP/lb? Wing area? How about wing loading?

If you'll note in the conclusion portion of the P-47D vs. P-47N zoom climb test report it reads...

"Above 5,000 feet the P-47N was superior.  This advantage in zoom can be mainly attributed to the higher power available of the P-47N."

"power available"=Thrust*Velocity portion of our Ps (specific excess power) equation.
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Offline Stoney

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Re: How to regain lost E? (F4U)
« Reply #61 on: August 06, 2011, 05:22:26 PM »
What excess power? HP/lb? Wing area? How about wing loading?

Stock P-47D-30 = 2200 HP on WEP with 300 ft^2 of wing area.

Stock P-47N      = 2800 HP on WEP with 330 ft^2 of wing area.  That's 600 HP and 10% more wing area = substantial performance difference.

Wing loading and power loading without context mean nothing.

But, specific excess power is a totally different thing.  See Tango's post above.
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Offline bozon

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Re: How to regain lost E? (F4U)
« Reply #62 on: August 06, 2011, 05:24:32 PM »
...
Ps = (thrust - drag) * velocity / weight
...
If you look at that relationship you can see that all else being equal (thrust, drag, velocity) increasing weight ALWAYS reduces Ps.
It always reduces Ps (in absolute value...), but not always the height of the zoom. Ps only tell you the rate of energy gain (per mass). It is not a direct measurement of the zoom. The benefit of the mass is implicit in this equation - it is in the time and the sign of Ps. Note that if Ps is negative, weight actually INCREASE Ps!

So, when is Ps negative? when the thrust is weaker than the drag. Most commonly this would be when the speed is high enough, so if the zoom is started from very high speed, the weight gives an advantage in the first part of the zoom - this is a ballistic-like phase (a stone thrown up into viscous air). Then as the speed drops, so does the drag and the thrust will eventually turn Ps positive - this is the climb phase, where engine pull significantly affects the dynamics and weight becomes a hindrance. Just remember that the drag dominates when the velocity is high, and we are multiplying by velocity to get Ps, so it becomes highly negative very fast. Thrust will dominate at low velocities, so if (thrust-drag) part increase, velocity typically decrease and Ps does not increase as much.

The time domain is a little more complicated. There is a third scale that affects the dynamics. Put in the term used in the Ps expression: "g" the gravity acceleration which competes with "thrust/mass" and "drag/mass". In WWII fighters, g is always the largest since the thrust is no where near as strong as gravity. This is why the differences between planes are not huge.

The biggest effect of some extra thrust/mass is how the zoom will look like close to its peak, rather then how high it will be (even though it will). The thrust, even if much weaker than gravity will make the last part of the zoom longer (in TIME) where the plane will appear to be "hanging on its prop". It is not really going anywhere (this is when the speed approaches 0), but it is not starting to fall either. Note that in the Ps expression, low velocity means small Ps and when it is close to zero it does not matter much how close it is. This is the phase where the plane gains the most from having low mass - but it is not much in terms of gaining alt because the velocity is low!

When the thrust starts to be a significant fraction of the weight, then things really start to change and mass makes a huge difference. Of course, for planes with thrust>weight, there is no proper "zoom" anymore - it is just a vertical climb.
« Last Edit: August 06, 2011, 05:28:15 PM by bozon »
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Offline 2bighorn

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Re: How to regain lost E? (F4U)
« Reply #63 on: August 06, 2011, 06:21:14 PM »
It's well known aerodynamically that the airplane with the greatest time average Ps in a zoom climb will out zoom one with a lower Ps.

What Stoney and you don't take into account (also well known) is state of energy (at zoom climb entry) and conservation of momentum.
Add that to your equation and it becomes halluva lot more complicated but also much closer to real life zoom climb tests.

Even today, with all the computing power, computational fluid dynamics models have to be confirmed by wind tunnel and test flight results, because in average they're 1.5-3% off. Simplified formulas for multitude of that. Why? Whilst in principle correct, they're extremely simplified.

There's a reason why fluid dynamics (and with that aerodynamics) is so damn complex.

Offline Stoney

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Re: How to regain lost E? (F4U)
« Reply #64 on: August 06, 2011, 06:54:08 PM »
What Stoney and you don't take into account

Tango, I told ya we forgot something!!!  I guess I need to get back in my books...   :noid
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Offline 2bighorn

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Re: How to regain lost E? (F4U)
« Reply #65 on: August 06, 2011, 07:21:33 PM »
This advantage in zoom can be mainly attributed to the higher power available of the P-47N."[/i][/size]

Of course, it had 18% more power at only 13% more weight.

Offline PuppetZ

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Re: How to regain lost E? (F4U)
« Reply #66 on: August 06, 2011, 08:37:58 PM »
If i get it right, the airplane with the highest thrust/weight ratio will out zoom the other if starting at the same speed? That's why a f-15 will outzoom a brewster  :devil
« Last Edit: August 06, 2011, 08:46:39 PM by PuppetZ »
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Offline dtango

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Re: How to regain lost E? (F4U)
« Reply #67 on: August 06, 2011, 08:43:21 PM »
If i get it right, the airplane with the highest thrust/weight ratio will out zoom the other if starting at the same speed? That's why a f-15 will outzoom a brewster  :devil

Yes, more or less.
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Offline pervert

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Re: How to regain lost E? (F4U)
« Reply #68 on: August 06, 2011, 08:44:10 PM »
Theres nothing more unsexy than fighter planes and equations  :old:

Offline dtango

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Re: How to regain lost E? (F4U)
« Reply #69 on: August 06, 2011, 08:47:41 PM »
Theres nothing more unsexy than fighter planes and equations  :old:

:)

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"At times it seems like people think they can chuck bunch of anecdotes into some converter which comes up with the flight model." (Wmaker)

Offline PuppetZ

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Re: How to regain lost E? (F4U)
« Reply #70 on: August 06, 2011, 08:49:40 PM »
:)

"Beware the lessons of a fighter pilot who would rather fly a slide rule than kick your ass!"  Commander Ron "Mugs" McKeown, USN, Commander of the USN FWS "Top Gun" School.

Sure, one got to have a grasp of these principle but when it come down to combat there is only brain and gut out there.
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Offline Puma44

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Re: How to regain lost E? (F4U)
« Reply #71 on: August 06, 2011, 09:04:10 PM »
If i get it right, the airplane with the highest thrust/weight ratio will out zoom the other if starting at the same speed? That's why a f-15 will outzoom a brewster  :devil

Zactly right!  :aok  You've got all the fomulas boiled down to the basic "pilot big picture".  An understanding of how to apply lift vector whilst maneuvering against an opponent is also very helpful.   :salute



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Offline dtango

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Re: How to regain lost E? (F4U)
« Reply #72 on: August 06, 2011, 09:08:43 PM »
bozon - I agree with most of what you said.  Two points I would make:

1) Ps = (T-D)*V/W = rate_of_climb + rate_of_flight_path_velocity_ change.  Yes, Ps is not a direct measure of zoom height.  However it includes a direct measure of rate of change of climb, thus the time average of Ps over the course of a zoom climb tells you which airplane is going to zoom higher.

EDIT:-------------
just a correction "rate_of_flight_path_velocity_ change" isn't the correct term.  Here's the equation:
Ps = (T-D)*V/W = dH/dt + (V/g*dV/dt)...
dV/dt is what I was referring to as "rate_of_flight_path_velocity_ change"
------------------

2) Mass absolutely is a factor which is why this topic totally throws people.  The question is how much and when.  Typically the portion of a zoom climb where T<D is very short in duration thus extra mass quickly becomes a detriment.  In the links I've posted above I've done some examples to demonstrate this (numerical integration over the time domain).  The fact that (for our prop aircraft) T increases with decreasing velocity while D decreases with decreasing velocity further ensures that T<D is only over a short duration.
« Last Edit: August 06, 2011, 11:52:14 PM by dtango »
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Offline dtango

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Re: How to regain lost E? (F4U)
« Reply #73 on: August 06, 2011, 09:15:31 PM »
What Stoney and you don't take into account (also well known) is state of energy (at zoom climb entry) and conservation of momentum.
Add that to your equation and it becomes halluva lot more complicated but also much closer to real life zoom climb tests.

Ah, my good sir but Ps absolutely takes momentum into account.  The derivation of Ps comes directly from F=ma.  The right-side m*a is the rate of change of momentum. Momentum is completely factored in.
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Offline dtango

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Re: How to regain lost E? (F4U)
« Reply #74 on: August 06, 2011, 09:27:09 PM »
Tango, I told ya we forgot something!!!  I guess I need to get back in my books...   :noid

What I forgot was to drink a few more frozen margaritas :cry.  Everything always magically becomes more rational after that  :D.
« Last Edit: August 06, 2011, 09:31:09 PM by dtango »
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"At times it seems like people think they can chuck bunch of anecdotes into some converter which comes up with the flight model." (Wmaker)