Expert: Paul Soderman - 6/3/2006

Question
Please I'm needing the best definition to "Best Angle" & "Best Rate" for jets and all kind off comments that you could bring to me, thank you very much

Answer
Hi Enrique

Here is something I wrote on best climb angle and best rate of climb for propellers.  I leave it to you to find the effects for jets.

The difference between propeller aircraft best (steepest) climb angle and best rate of climb can be confusing.  But if we break it down step-by-step I think it can be made clear.  But first, we have to define some terms.

You are right, Thrust is a Force:  F = ma
In steady level flight, thrust equals drag, so the net force is zero and acceleration is zero.

And you are right again; power is the rate of doing work.  Rather than deal with engine power and try to figure efficiencies, for this problem it is best to think of the power developed by the propeller, which is total thrust times aircraft airspeed:

P = TV                           (1)

To climb, an aircraft must develop excess thrust Tx, which is simply the amount of thrust in excess of drag and the weight component in the direction of climb.

Excess power is then excess thrust time aircraft airspeed:  

Px = TxV                        (2)

Consider a vector diagram of the velocities in climb and the key forces in climb.  

c = climb angle
Vtas = true airspeed along flight path
Vg = ground speed
Vv = vertical speed
Tx = excess thrust
W = aircraft weight
L = aircraft lift

   

It is clear that the climb angle is just:     c = sin-1 Vv/Vtas         (3)

And,                  c = sin-1 Tx/W            (4)

This is all intuitive, maximize thrust and the aircraft will achieve the best climb angle at some airspeed Vx.  As we gain altitude, the power available and excess thrust decreases because air density and pressure are decreasing.  Even a turbocharger cannot ram enough air into the engine to maintain power.  However, the aircraft accelerates somewhat because drag is going down with density also.  Just as you said, the airspeed for best climb angle increases from sea level to the service ceiling, which is the point where excess thrust is zero and climb is no longer possible.  Notice from  the above sketches, that the climb angle is actually decreasing as we climb because Tx decreases, but it is still the optimum climb trajectory if thrust is maximum.

Assume for a moment that you started up the best climb angle, but really wanted to achieve the best rate of climb.  If you pushed the nose down, the aircraft would accelerate because the weight component in the flight direction would decrease.  The best rate of climb occurs at airspeed Vy when the aircraft maximizes the excess power available.  This can be shown from equations 3 and 4 if we solve for the vertical velocity Vv, which is what we are trying to maximize:

  Vv = Tx Vtas / W = Px / W                  (5)

As before, propeller power available and excess power decrease with altitude.  But here comes the tricky part.  The airspeed for the peak power available actually increases with altitude.  But it increases only slightly.  Because of the density error at altitude inherent in airspeed indicators (they read low), the indicated airspeed Vy will decrease with altitude.  In any case, the airspeed for best rate of climb, Vy, always converges to the airspeed for best climb angle, Vx, at the service ceiling.  

Hope this helps.

Paul