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3 Answers

Excess Thurst versus Excess Power

Asked by: 23082 views Aerodynamics

I can read and remember what the book says (I think) about excess thrust determines rate of climb and excess power determines angle of climb. But I dont have a good way of understanding this (like a car and hill analogy or something). I want to know more than just the memorized answer, is there a good analogy or way of understanding this?

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3 Answers



  1. Steve Pomroy on Feb 11, 2011

    Hi Matthew.
     
    First of all, excess thrust determines your angle of climb, and excess power determines your rate of climb, not the other way around:)!
     
    Let’s look at what thrust and power are.  That’s probably the best (and most overlooked) starting point.
     
    Thrust is a force, as in one of those things Sir Isaac talks about.  An applied force, if unbalanced, will make an object accelerate.
     
    Power is not a force, although it is related to force, and a force will be present when there is power.  We’ll come back to what exactly power is in a moment, but first, two other definitions:  work and energy.
     
    Work is the transfer of energy from one object to another.  It is calculated as a force applied over a distance.  Energy is an object’s ability to do work.  Energy comes in the form of kinetic energy (speed) and potential energy (height or chemical energy in our fuel).
     
    So, what is power?  Power is the rate of doing work — the rate of transfering energy from one object to another (you might note now that this will eventually lead to rate of climb, where our rate of climb corresponds to our rate of gaining potential energy).
     
    Forces are vectors.  In order for them to balance, we need to account for their direction.  If we are producing more thrust than drag, the aircraft will accelerate — or we can balance the excess thrust with a component of weight by inclining the thrust upward.  The angle we need to incline it will depend on how much excess there is compared to how much weight we have.
     
    Power is not a vector, it’s a scalar.  So angles don’t come into consideration when dealing with power.  Rates do.  The engine provides the aircraft with energy at some rate.  The airstream robs the aircraft of energy at some rate (via drag).  If the engine is providing excess power, the aircraft will either accelerate (gain kinetic energy) or climb (gain potential energy).  The rate at which we accelerate or climb will depend on the amount of excess power compared to our weight.
     
    To take an example in which the distinction between force and power may be a bit more intuitive think about riding a bike.  More specifically, think about riding a bike up a hill:
     
    The stepper the hill, the more force you’ll need to apply to the pedal to keep moving at a constant speed.  In other words, more force (i.e. – thrust) corresponds to a steeper hill (i.e. – angle of climb).  The rate at which you climb, however, depends not only on the force you apply to the pedal, but on how fast you move the pedal (how many pedal strokes per minute).  Moving at a constant speed — no matter what that speed is (if we ignore for simplicity the changes in drag force) — you will always be applying the same force to the pedal.  But at higher speeds (and therefore higher climb rates) you will be excerting more power because of the increase in the rate of pedal movement.
     
    So, bottom line here:  Force (thrust) corresponds to climb angle, and power corresponds to rate of climb.
     
    As you can see, the force v. power distinction is really not that simple.  I’m afraid simplifying it further than this discussion presents the risk of being inaccurate.  In any case, I hope this helps!
     
    Cheers,
    Steve
    http://www.flightwriter.com

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  2. Wesley Beard on Feb 11, 2011

    Matthew, I think of it like this.  Power = Thrust / Time.  The only one that has a time listed is Vy or best rate of climb and so excess power is used to calculate what Vy is.  The other must be Vx or best angle of climb and excess thrust is used in that calculation.
     
    Steve’s answer is quite good. +1

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  3. Matthew Beyer on Feb 11, 2011

    Thanks Team, that was a perfect answer. I like the depth. I printed it and read it many times to digest it. Then the analogy is icing on the cake. Thanks

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