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Power vs Thrust (again)

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Power vs Thrust (again)

Old 9th Apr 2013, 16:51
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Power vs Thrust (again)


Yes yes, another Power vs Thrust question
I understand the basics, I know the details, but I don't really understand the details.
I searched through the forum, read all the power vs thrust threads, but still I'm struggling with it.
Here is what I know (please correct me if I'm wrong):
  • Power = thrust x time
  • Piston engine: engine produces power, fuel flow depends on power produced, power goes to propeller, propeller produces thrust.
  • Jet engine: engine produces thrust directly, so fuel flow depends on the thrust produced.
Those 3 arguments are perfectly clear.
  • All engines: best angle of climb speed (Vx) happens when the excess thrust is the greatest
  • All engines: best rate of climb speed (Vy) happens when the excess power is the greatest
It's with those 2 I have an issue.
For Vx I'd say okay, excess thrust. Why? A jet engine produces thrust, so here we're already talking about thrust. And with a piston engine: it's not because there's a lot of excess power that there's the same amount of excess thrust (a propeller's effiency depends on a lot of other factors).
But what about Vy? Why can't we say that for the best Vy we need the greatest excess thrust? Okay, power = thrust x time, and with Vy time is an important factor. But: thrust = power/time. Here we also consider the time.

Hopefully I'm clear about what it is that I don't fully understand.

Thanks in advance.
Spir4 is offline  
Old 9th Apr 2013, 17:02
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Might be some confusion over the word Thrust in
Power = Thrust x Time
and the word Thrust in the Vx equation

If an engine produces 250HP of power for 5 hours with no propeller, how much thrust did it have? none? how can the equation be true then?

In the first equation, Power = Thrust x Time is not thrust of the airplane through the air, but thrust of the engine.

In the second equation, Thrust is thrust through the air (propeller thrust), not thrust of the engine.

Thrust through the air, or propeller thrust, is related to the efficiency of the propeller, as well as the power of the engine.
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Old 9th Apr 2013, 18:15
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Okay I get half of that

Why talk about thrust of an engine when the thrust is not related to motion (through the air in this case)?
Without that motion, if you talk about thrust of an engine, isn't it like talking about the apples of an engine? Kind of a non-related word.
An (pison) engine produces power, which turns the propeller, which produces thrust.

Still not clear I'm afraid
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Old 9th Apr 2013, 18:23
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First off, power does not equal thrust multiplied by time. Below are correct equations.

Work = force x distance
Power = work / time
Power = torque x angular velocity (e.g., RPM)
Torque = force x length of lever arm

Power = force x distance / time
Velocity = distance / time


Power = force x velocity


Power = thrust x TAS

You first need to understand all those equations. If you're unsure about one of them ask for clarification. This page might help: http://en.wikipedia.org/wiki/Power_(physics)

Regarding piston engines:

The expansion of gas inside the cylinder creates a force on the piston and moves it a certain distance. Force x distance = work, meaning the burning of the mixture inside the cylinder does work on the piston. Since power = work / time, the shorter the time period (i.e., faster) the work is done in, the more power is required.

That power gets transferred into torque as the connecting rod applies a force a certain distance from the center of the crankshaft. So torque is now being applied to the crankshaft. Torque overcomes the friction and begins moving the crankshaft at a certain speed (angular velocity [RPM]).

To calculate the power created at the crankshaft you take the torque applied to it and multiply it by the RPM of the crankshaft.

The propeller then converts that power into thrust. The power from the engine is required to overcome the drag due to lift (thrust in this case), parasitic drag, and internal friction. When RPM is constant - the propeller-drag torque, plus internal friction, is equal to the torque applied to the crankshaft by the expanding gasses in the cylinder. Basically, the moments (torques) clockwise about the crankshaft center equal the moments (torques) counterclockwise about the crankshaft center.

Regarding climbs:

Forget about Vx and Vy for a moment - every climb has a certain climb angle and climb rate! When you're thinking about excess thrust and excess power just think of any sort of climb - not Vx or Vy. What it's saying is that the angle of climb (regardless of what it is) is directly related to the excess thrust, and rate of climb (regardless of what it is) is directly related to excess power.

These two images are from Aerodynamics for Naval Aviators: Climbing - Imgur

For the angle of climb bit, it's a simple analysis of the 4 forces in flight. At the top of Figure 2.21 you'll see how the forces are broken up for analysis. Essentially the more excess thrust you have, the more horizontal component of weight you're able to balance, meaning the steeper you can climb. Reading those pages I've attached should explain it quite clearly.

As for the rate of climb bit, you'll see at the top of Figure 2.21, beside the airplane, a triangle where rate of climb, velocity, and angle of climb are plotted. Essentially, for the same angle of climb, increasing the TAS (velocity) will increase the rate of climb directly. Since thrust is directly related to angle of climb, power = thrust x TAS. That's explained in more detail on the second image above.

In a combustion engine, the power is related to the speed of the engine (RPM). In the case of the airplane, power is related to the speed of the airplane. There is no difference in this relationship when comparing piston to jet performance. In the case of a propeller engine, the power is called "brake or shaft horsepower"; for an airplane, the power is called "thrust power" or typically "thrust horsepower" (THP) as it's generally displayed in units of horsepower. Piston, turboprop, and jet airplanes all have THP and in all 3 cases it's the same - thrust x TAS.

The only difference happens because there is a difference in how a prop engine's thrust varies with TAS to how a jet's thrust varies with TAS. Since a jet's thrust is relatively constant throughout the range of TAS, the power (THP) will increase linearly with an increase in TAS. Notice that ALL THP vs TAS graphs will start at 0,0. That is because at zero TAS, the THP is zero! Even though you're at the runway with the engine at full BHP/thrust, the velocity of the airplane relative to the air is zero, therefore, the THP is zero. It's that simple! I'm ignoring wind. As for the prop engines, the thrust will decrease with increasing TAS because prop efficiency goes down. Initially the decrease in prop efficiency (thrust) is not enough to overcome the increasing TAS, so THP will increase. Eventually, the decrease in thrust overcomes the increase in TAS and you'll notice the THP curve will taper to zero and then decrease.

Now going back to Vx and Vy:

If you want the maximum angle of climb, you need the maximum excess thrust. If you want the maximum rate of climb, you need the maximum excess power.

Additional reading:

US Navy - Thrust and Power

Aerodynamics for Naval Aviators - Propeller Efficiency

Last edited by italia458; 9th Apr 2013 at 18:51.
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Old 9th Apr 2013, 18:42
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Power = thrust x time
Those 3 arguments are perfectly clear.
I argue they are not...

thrust x time = impulse, and probably not relevant to this discussion

thrust is a force created by the propulsion system

work = force x distance = energy delivered by the propulsion system

= energy rate = energy / time

power generated
= thrust x distance / time = thrust x velocity

power consumed
= fuel flow x specific energy

efficiency = power generated / power consumed

Now consider an aircraft in climb.
Climbing, in energy terms means increasing potential energy.

potential energy = weight x height
climb speed = change of height / time

The highest climb speed occurs when you have the highest excess power, i.e. the highest rate of adding energy.

When looking at the climb angle, we abandon the energy model and look at forces only.

Along path there are three forces acting on the aircraft.
1. Thrust
2. Drag
3. Weight x sin(climb angle)

For constant groundspeed climb, the forces are in equilibrium.

Thrust acts opposite to Drag and the along path Weight component
The higher the difference between thrust and drag, the great we can make the climb angle.
The maximum climb angle is achieved at the speed where maximum thrust - drag is maximal.

Note that typically you will fly a constant CAS or constant Mach climb which requires change of groundspeed during the climb and therefore the forces will not be in equilibrium.
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Old 9th Apr 2013, 18:43
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Here is another reference. It's from the JAA ATPL Performance book.

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Old 9th Apr 2013, 19:09
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But what about Vy? Why can't we say that for the best Vy we need the greatest excess thrust? Okay, power = thrust x time, and with Vy time is an important factor. But: thrust = power/time. Here we also consider the time.
As Italia has said, your equations (power = thrust x time) and (thrust = power / time) are both incorrect.

To understand when best ROC occurs when excess power is greatest it might be better to use the equations below.

Power = work done / time taken to do that work.


Power = force x velocity

In a climb work must be done to achieve the following two things:

1. Move the aircraft forward at whatever TAS we have chosen. To achieve this we must exert a thrust force that is equal to the drag. So the power used to carry out this part of the work is equal to drag x TAS. This is the POWER REQUIRED simply to maintain our TAS.

2. Move the aircraft vertically upwards at whatever vertical speed (ROC) we have chosen. To achieve this we must apply a vertical component of thrust that is equal to the weight of the aircraft. So the power that is required to achieve this = Weight x ROC.

The power in paragraph 2 (Weight x ROC) is the extra power (or excess power) that we need to achieve the selected ROC. If our maximum power available is just equal to the power required (in paragraph 1) , we can maintain our selected TAS, but we have no extra power available to achieve any ROC. So to achieve the maximum possible ROC we must fly at the speed at which we have the maximum amount of excess power available.
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Old 9th Apr 2013, 19:50
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@ italia458: Thank you a thousand times. This is an issue that has been bothering me since my PPL lessons. I just accepted everything, because nobody I've met so far was able to give me a clear answer. You just did, and you made my day !

@ ATCast and keith williams: Also a big thanks for your explanations !

With those clear explanations, it all sounds so simpel and logically. I don't know why I never figured out the correct explanation. I guess I combined different source materials the wrong way.

Thanks again !!!
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Old 9th Apr 2013, 21:55
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I'm glad I could help! I've been in your position before regarding this exact subject and know the feeling.
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Old 10th Apr 2013, 05:03
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Not to detract from what was written before as it sounds familiar-ish from way back.

Can a simplified visual model be related to this?

Simply put the aircrafts polar curve, min. sink (best endurance) ~Vx and L/D max. (best range) being always a higher IAS/TAS than Vx although usually Vy is faster than L/D max. to accomodate other issues, cooling, thrust/power curve etc.

Gets a bit messy linking turbine best range speed with L/D max due to the nasty habit turbines have of consuming 1/2 their fuel just to keep running.

If I've got this wrong I hope someone will straighten me up.
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