Performance of the Fw 190A on the Deck?

[Harri Pihl]

Originally Posted by drgondog

Hari - Thrust is a Force not Power.

Yes, and that is what I have been saying all the time.

What I do is that I convert the power (W) to the force (T) according to propeller efficiency (n) and speed (V):

T=(n*W)/V

And your failure is that you don't understand that this means that the force (thrust) varies with speed at constant power. Lets quote the Hamilton Standard Red book:

[Franek Grabowski]

Bill

Franek - you are right about yaw issues in dives from .75 to .83
But this discussion is about level deck speeds in the .55 to .6 range.
In a dive you have to feed rudder to keep the nose from wandering but it was not a divergence issue as 'highly unstable' which it wasn't.

I just wanted to point out that there were some aerodynamical differencies between the two types. Certainly the wetted area was different for both, especially on the fuselage.Of course, the question is, how it affected the aircraft performances, being in turbulent airflow from propwash. (I hope it is still in English)

Aero engineering is a complex field. Getting precision estimates of performance, pre flight or wind tunnel testing for a new airframe and geometry was a combination of arcane combinations of empirical factors (i.e "e" and "propeller efficiency" and even various components of parasite drag like wheel wells, new antennae, bomb racks, etc.).

Yes, indeed, but we often omit several such components, like a parasite drag of riveting and panel lines, sometimes multiplying results by a specific factor. Then we have a interference drag, which we may get approximate or measure on wind tunnel test, but it is rather hard to predict. We can also simplify the model and do not calculate drag of antennae, etc. either omitting it or adding a constant figure, depending on influence on a total drag.

Having said this, this discussion and Crumpp's presentation and arguements are about Parametric studies on the same airframes with one variable changed.. so we don't have to screw with "e", AR, predictions about flow separation, etc.

Yes, indeed, but the problem is with estimating influence of one variable - change of weight caused by fuel consumption - on overall aircraft performance. I tend to agree with Graham, based on my aerodynamical knowledge with very limited practice, that it is minor.

On the other hand when flight testing the same airframe, with low fuel versus full combat load - same engine, same airframe, same boost, same pilot, same geography and easy to calculate variations for Temp and Altitude

Well, we will never enter the same river to be exact. But of course we can use an approximate and do the comparison. Let's say the pilot is able to put an aircraft into same conditions in air but weight. The question is, how we would measure change of weight of flying aircraft, and how we will calculate airspeed? I think accuracy of measurements is a key problem.

The P-51. For the answer to the obvious question compare the Drag of each airfoil and airframe to get the "aha". Speed and range.

Oh yes, indeed. Heavier aircraft, powered by the same engine, but with better aerodynamics is substantially faster. On the other hand, the lighter one, Spitfire, is generally a better climber and able to fly higher than any Mustang. Different parameters should be considered.

Franek - I respect Graham's knowledge but Crumpp in my opinion based on my own academics and practice is entirely correct.

Yes and no. Graham in my opinion correctly pointed out that the change of weight caused by fuel consumption of an average WWII fighter is just too small to be considered important factor while discussing the maximum speed. Theory is generally correct but it has some limitations, especially if we discuss profile of the mission.

The question for the example discussed is ~ 10mph significant for the 6% increase in weight? It would be for me, particularly if my Mustang was in the lower range of performance in the production series..

I bet it would be much more important for you, to have it smoothly polished, with no dents, and to have fuselage tank empty in case of dogfight, to avoid stability problems. Let's not forget one thing as well. We discuss actual performance of aircraft in combat. talking about maximum speed of Mustang, let's not forget nobody would enter dogfight with full fuselage tank and wing tanks. Mustang entering combat over Normandy and over Berlin had similar weight and it would not affect your performance. You would not fly to both locations at your maximum speed as well.

That said, I am curious of the mentioned performance comparisons of Mustang, and wondering if it is the same thing I have seen some 10 years ago. No comments, till I see the stuff.

[Crumpp]

Nope, in that case I would have thrust watts but I did not calculate that out from the calculation.

I said you wouldn't understand a dimensional analysis. It is pretty obvious that a discussion with you cannot go anywhere. Don't mistake a lack of response with being correct.

Bill was refering to the muliple times you have confused them over the course of the thread Harri. Not a carefully worded response your Kabuki theater.

Yes and no. Graham in my opinion correctly pointed out that the change of weight caused by fuel consumption of an average WWII fighter is just too small to be considered important factor while discussing the maximum speed. Theory is generally correct but it has some limitations, especially if we discuss profile of the mission.

That theory is very much like saying a small cancer tumor is not important.

Not one aerodynamic text can be found that states weight is unimportant because one of the affects is a relatively small reduction in top level speed.

Obviously references are available. One has to wonder why there are not used.

Lets quote the Hamilton Standard Red book:

All the best,

Crumpp

[Harri Pihl]

Originally Posted by Crumpp

I said you wouldn't understand a dimensional analysis.

The dimensional part of the calculation is, after initial conversions, entirely at SI units. There is no THP calculated, nor anything equivalent like "thrust watts". The power is directly converted to thrust according to n and V.

If someone clearly has problems with dimensional units here, he is the one who wrote that:

"When you convert that Horsepower to SI units and then mulitply by efficiency...

You have thrust horsepower!"

You wrote that.

Originally Posted by Crumpp

Bill was refering to the muliple times you have confused them over the course of the thread Harri. Not a carefully worded response your Kabuki theater.
...
Obviously references are available. One has to wonder why there are not used.

Well, then we have an interesting situation here.

On one side we have drgondog, supported by you, who claim that thrust remains constant.

On other side we have Hamilton Standard (designer and builder of the propeller for the Mustang) who claims that thrust varies with speed just like I did in the calculation.

[drgondog]

Let's take one more pass at this.
Assume max TO power, brakes on, zero velocity. I say that is the condition of Max Thrust Available for the system at that altitude.

The system does not accelerate, no drag forces are experienced and the force retarding the thrust is the friction on the wheels with the brakes creating the torque to keep the wheels from rotating.

Release brakes, Same Max Thrust Available.

The Max Thrust available exceeds the thrust required and the system accelerates until the speed is sufficient to create enough lift to overcome weight. Same Max Thrust but the a/c is airborne and continues to accelerate.

Drag forces increase as the system accelerates until the drag forces equal the Max Thrust Available... at that point the system has reached maximum velocity for that system with that Max Thrust available. It is in equilibrium.

At no time did Max Thrust available change from V=0 through V=Vmax for that weight condition.

Add 500 pounds of fuel.

Everything is the same for this System B except that induced drag increases throughout the profile in comparison to the lighter weight system

MAX THRUST available for the System B is the same as it was in the lighter weight system, but the aircraft will achieve equilibrium at a lower speed because the Thrust required to attain an equal velocity of the lighter weight system is insufficient to overcome the increased induced drag due to the heavier weight system.

Summary

Max Thrust for System A = Max Thrust for Sytem B
Max Weight for System A < Max Weight for System B
Max Velocity for System A > Max Velocity for System B
Max Thrust for System A still = Max Thrust for System B

Think of it another way

System A pops a drogue chute at max speed and Max Thrust
Drag increases dramatically, speed drops dramatically, induced drag drops dramatically, parasite drag increases astronomically,

System A achieves equilibrium at lower speed because the Max Thrust Avaialable is lower than Thrust required to maintain the higher speed.

System A makes no changes to internal weight, makes no changes to power settings or altitude, but slows down dramatically...

So, Harri - by your use of the conversion of Jet Thrust to THP conversion equations do you believe that Thrust just increased in the example above by virtue of the reduction in velocity as described in System A above?

The answer is no.

The Max Thrust Available did not change from zero Velocity to Max velocity, nor did the weight of the system change. Only the parasite drag increased and induced drag decreased causing the System to decelerate to a lower velocity. No Thrust change, no Hp change

That equation is all about converting the thrust of a jet to an estimate of work being performed on a system by a propeller/piston engine system, and with modification, to a turbo prop.

I don't have the Hamilton Standard 'Red Book', but it probably has something similar to page 109 of the Blue Book - THRUST-HORSEPOWER CONVERSION in the GAS TURBINE section.

Ask yourself how real Force increases with a reduction in velocity. If you looked at the two states of System A, the first at rest with brakes applied and the second at Vmax - both with same HP/propeller combo/rpm, by your anology the Thrust would be infinite when the system is not moving?

[Franek Grabowski]

Originally Posted by Crumpp

That theory is very much like saying a small cancer tumor is not important.

The comparison is not valid. You can refer to cancer discussing corrosion, but not speed. Simply if you have few pounds more or less on your body it will not affect your performance. If you are 100 lbs overweight, then it will. I consider that due to low speed we do not discuss aerodynamics of human body.

Not one aerodynamic text can be found that states weight is unimportant because one of the affects is a relatively small reduction in top level speed.

What for? Everyone who calculated performances will know.

Obviously references are available. One has to wonder why there are not used.

Use your references and show us on an equation a full formula for P-51 speed. Please explain us what percentage of speed is affected by what percentage of factor.

[Harri Pihl]

Originally Posted by drgondog

Let's take one more pass at this.
Assume max TO power, brakes on, zero velocity. I say that is the condition of Max Thrust Available for the system at that altitude.

The system does not accelerate, no drag forces are experienced and the force retarding the thrust is the friction on the wheels with the brakes creating the torque to keep the wheels from rotating.

Release brakes, Same Max Thrust Available.

At the beginning speed is zero (V=0) so the formula:

T = (n*W)/V

can't be used but as example the one for V=0 in the Hamilton Standard book (above formula for V>0).

Originally Posted by drgondog

The Max Thrust available exceeds the thrust required and the system accelerates until the speed is sufficient to create enough lift to overcome weight. Same Max Thrust but the a/c is airborne and continues to accelerate.

This is mostly true for the jets but not for the propeller planes. We can assume that power remains fairly constant (save the RAM and altitude effects), but thrust will vary according to formula for V>0.

Originally Posted by drgondog

Drag forces increase as the system accelerates until the drag forces equal the Max Thrust Available... at that point the system has reached maximum velocity for that system with that Max Thrust available. It is in equilibrium.

At no time did Max Thrust available change from V=0 through V=Vmax for that weight condition.

This part is only partially true, the plane will reach equilibrium but while the velocity increase, the thrust at constant power will decrease.

Originally Posted by drgondog

Add 500 pounds of fuel.

Everything is the same for this System B except that induced drag increases throughout the profile in comparison to the lighter weight system

MAX THRUST available for the System B is the same as it was in the lighter weight system, but the aircraft will achieve equilibrium at a lower speed because the Thrust required to attain an equal velocity of the lighter weight system is insufficient to overcome the increased induced drag due to the heavier weight system.

This part is also only partially true because thrust increase when the speed decrease.

The power is constant, however, the thrust depends on speed.

IMHO the main problem here is that you have apparently played with the jets.

Originally Posted by drgondog

So, Harri - by your use of the conversion of Jet Thrust to THP conversion equations do you believe that Thrust just increased in the example above by virtue of the reduction in velocity as described in System A above?

It's not just my use of this very basic formula but also Hamilton Standard, Hoerner, NACA, NASA, RAE... you name it. Below is quote from Hoerner.



Note that the speed value is in foots per second and 1200 is amount of power in hp.

[drgondog]

[quote=Harri Pihl;70657]At the beginning speed is zero (V=0) so the formula:

T = (n*W)/V

can't be used but as example the one for V=0 in the Hamilton Standard book (above formula for V>0).

So, at 1kt/hr Thrust is near maximum?, at 2Kts the thrust is coming down and at 300kts it is at minimum?


This is mostly true for the jets but not for the propeller planes. We can assume that power remains fairly constant (save the RAM and altitude effects), but thrust will vary according to formula for V>0.

Power is basically constant, Thrust (Force along positive horizontal axis exerted on the stsyem by the engine/propeller combination) is basically constant for this system from V=0 kts and V= 300kts and every velocity in between.

It does not start at 1kt (or 10 or 100kt) at maximum and decrease with Velocity. If you think so, tell me at what velocity you want to start using the THp formula?



This part is only partially true, the plane will reach equilibrium but while the velocity increase, the thrust at constant power will decrease.



This part is also only partially true because thrust increase when the speed decrease.

The power is constant, however, the thrust depends on speed.

IMHO the main problem here is that you have apparently played with the jets.

Actually true, and I mostly played with engineering and physics.

Force = Thrust = d/dt(MV) ..

in this system (through a propeller plane) the thrust is the net mass flow rate throught the prop plane. Unlike a jet engine it has no additional combustion energy converted to the system as thrust all the engine torque must be converted through the propeller system to impart energy onto the stream tube entering the prop plane where it dramatically accelerates the mass flow rate of the stream tuve from Vfreestream to Vpropstream.

I am discounting exhaust thrust from this discussion as it is negligible to the thrust of the engine prop system until you get to the critical altitudes of the airframe/engine system.

It's not just my use of this very basic formula but also Hamilton Standard, Hoerner, NACA, NASA, RAE... you name it. Below is quote from Hoerner.



Harri - I know the equations. Look at the polar diagrams of the same ship (c-54, B-17) etc that shows the the effect of added weight to the same system and engine capabilities on cruise speed and range.

quote]

Thp is different from Thrust. Thrust of a system is required to do a free body diagram on the system to understand the result of variations in drag due to Weight.

The equations are about equating 'push' of a jet to transmitted Hp from an engine generating torque on a propeller. At 375 mph a system that generates one hpxprop efficiency= equates to 1 pound of thrust = 1 Thp.

Forget about Thp, Hp, and Thrust of a jet.

Focus on Force(s) retarding acceleration versus Force causing acceleration until equilibrium is attained. This is what I say doesn't decrease as a function of Velocity.

It is the same at zero mph, 20mph, 200mph and 340mph for that P-51B-15 with a 1650-7 at 67"/3000 rpm at sea level. The difference is that the Force > Drag until equilibrium is achieved.

If you wish we can agree to disagree - I can live with that.

[Holtzauge]

Originally Posted by drgondog

Let's take one more pass at this.
Assume max TO power, brakes on, zero velocity. I say that is the condition of Max Thrust Available for the system at that altitude.

The system does not accelerate, no drag forces are experienced and the force retarding the thrust is the friction on the wheels with the brakes creating the torque to keep the wheels from rotating.

Release brakes, Same Max Thrust Available.

The Max Thrust available exceeds the thrust required and the system accelerates until the speed is sufficient to create enough lift to overcome weight. Same Max Thrust but the a/c is airborne and continues to accelerate.

Drag forces increase as the system accelerates until the drag forces equal the Max Thrust Available... at that point the system has reached maximum velocity for that system with that Max Thrust available. It is in equilibrium.

At no time did Max Thrust available change from V=0 through V=Vmax for that weight condition.

Add 500 pounds of fuel.

Everything is the same for this System B except that induced drag increases throughout the profile in comparison to the lighter weight system

MAX THRUST available for the System B is the same as it was in the lighter weight system, but the aircraft will achieve equilibrium at a lower speed because the Thrust required to attain an equal velocity of the lighter weight system is insufficient to overcome the increased induced drag due to the heavier weight system.

Summary

Max Thrust for System A = Max Thrust for Sytem B
Max Weight for System A < Max Weight for System B
Max Velocity for System A > Max Velocity for System B
Max Thrust for System A still = Max Thrust for System B

Think of it another way

System A pops a drogue chute at max speed and Max Thrust
Drag increases dramatically, speed drops dramatically, induced drag drops dramatically, parasite drag increases astronomically,

System A achieves equilibrium at lower speed because the Max Thrust Avaialable is lower than Thrust required to maintain the higher speed.

System A makes no changes to internal weight, makes no changes to power settings or altitude, but slows down dramatically...

So, Harri - by your use of the conversion of Jet Thrust to THP conversion equations do you believe that Thrust just increased in the example above by virtue of the reduction in velocity as described in System A above?

The answer is no.

The Max Thrust Available did not change from zero Velocity to Max velocity, nor did the weight of the system change. Only the parasite drag increased and induced drag decreased causing the System to decelerate to a lower velocity. No Thrust change, no Hp change

That equation is all about converting the thrust of a jet to an estimate of work being performed on a system by a propeller/piston engine system, and with modification, to a turbo prop.

I don't have the Hamilton Standard 'Red Book', but it probably has something similar to page 109 of the Blue Book - THRUST-HORSEPOWER CONVERSION in the GAS TURBINE section.

Ask yourself how real Force increases with a reduction in velocity. If you looked at the two states of System A, the first at rest with brakes applied and the second at Vmax - both with same HP/propeller combo/rpm, by your anology the Thrust would be infinite when the system is not moving?

Drgondog, You seem to have problems understanding the propeller thrust formula Harri is using and it's limitations. In the context that has been discussed here (max speed) it is most certainly valid. However, when the plane just starts moving at very low speeds as in your example it is NOT valid. Since you claim to have an Ms in aeronautical engineereing I'm frankly surprised you do not know this and imply the formula is wrong since the thrust would be infinite at zero speed. Look up any old NACA report dealing with propeller efficiency and you will see that the propeller efficiency goes down to zero at zero speed. So we end up multiplying something approaching zero with something approaching infinity. However, at any speed that is interesting for performance analysis, e.g climb, turn and speed performance the formula is perfectly valid.

All the talk in this thread about the meaning and definition of "thrust horsepower" is as meaningful as discussing how many angels can dance on the tip of a pin. You can use different methods to arrive at a correct answer and I would certainly like to be shown why it is not possible to do the analysis according to the method provided by Harri and why one MUST use "thrust horsepower" or else is doomed to failure.

Using the thrust formula Harri posted above in the analysis has a clear advantage over the power method: For a WW2 figther, there is a significant part of the thrust (especially at high speed) that is derived from exhaust thrust. The exhaust thrust is close to constant with speed. It would be interesting to hear how you account for this in the "thrust horsepower" model.

[Holtzauge]

Originally Posted by drgondog

I am discounting exhaust thrust from this discussion as it is negligible to the thrust of the engine prop system until you get to the critical altitudes of the airframe/engine system.

Well drgondog, If you do have that aeronautical degree you mentioned before then you obviously need to do a refresher course: For a WW2 fighter, the ballpark percentage of the total thrust at top speed from exhaust thrust is 10-15%