# Is "excess thrust" possible in a steady state vector diagram?

Being recently very much sold on the lift-weight-thrust closed triangle (with weight forming the hypotenuse) model for climbing flight, I proceeded to put the logic through its paces (validation) and got stuck on one fairly basic concept: excess thrust may be only transient.

Increasing throttle increases speed. We can initiate a climb with more speed, thrust working against increased drag, and to hold the climb against the tilted lift vector by increasing the lift coefficient, by way of increasing AOA.

Note the with a GA aircraft (25% max thrust to lift), increasing AOA excessively leads to loss of airspeed. In fact, this is how to stall an aircraft.

Pitch can never be transitioned into hover before stalling.

Can anyone confirm that, beyond applying throttle before the takeoff roll, "excess thrust" exists at all?

• "Excess thrust" means "more thrust than required for level flight", not "more thrust than required for unaccelerated flight". I think you may just be stuck on this definition? Nov 12 '20 at 12:16
• @Sanchises that would be excess thrust available. Not meaning to be a trouble maker, but that "lift-drag-weight" vector diagram is in question (for light GA aircraft). Nov 12 '20 at 15:02
• I'll post an answer later. The free body diagram cannot be in question for it would spell the end of Newtonian physics :) Nov 12 '20 at 15:48
• Yes, excess thrust always exists when the airplane is accelerating. But that is not strictly steady state. However, it is perfectly possible to extend the vector triangle by an acceleration term which balances the extra thrust. Same with climb: This is not strictly steady state but can be modelled quasi-stationary because parameters change only slowly. In the end, even unaccelerated horizontal flight is not strictly steady state since fuel is burned off, so the aircraft mass declines over time. Nov 12 '20 at 17:49
• @RobertDiGiovanni: In the first moment speed is still slow and all drag increase is from the stronger slipstream. In a climb thrust must be higher because weight has a backward component (relative to the thrust line) and some of the weight is carried by engine thrust. The same reason why lift becomes smaller in a climb: The engine takes over a bit. Nov 12 '20 at 19:54

First, we must be clear on what exactly is "excess thrust". I will list two possible definitions, although many more may be possible.

1. Excess thrust is the component of the resultant force in the direction of the flight path. Using this definition, excess thrust and steady state flight are directly at odds (because any net force results in an acceleration). So in that case, no, excess thrust is not possible in steady state flight.
2. Excess thrust is thrust minus drag. It splits the big lumpy thing called 'thrust' in two numbers: one that is used to counter drag (which is always there), and another that can be used for useful things.

Since we already answered the first option, I will now produce the free-body diagrams for the second option.

On the left, we have level flight. We recognize the four fundamental forces of flight: lift (blue), weight (green), drag (red), thrust (orange). It is immediately clear that all forces cancel each other, and we will happily coast along.

In the middle, we have a steady state climb. All vectors, except the weight vector, are 'tilted' back and change magnitude. Lift has decreased slightly, and as a consequence, drag has also decreased minutely (although this doesn't really matter). The biggest change can be seen in thrust, which has almost doubled in magnitude!

On the right image, I have proven that despite all these changes, the net force is still zero. I have replaced the lift- and drag vector in purple by subtracting drag from thrust. But wait a second! That purple vector is exactly what I called 'excess thrust': thrust minus drag! So this purple vector gives us a nice indication of how much 'extra' thrust we are using to make sure the resultant force on the airplane is still zero.

The sometimes heard term, "max excess thrust available" is then an indication of how far we can 'tilt back' the blue lift vector. On the diagram on the right, it should be obvious that this 'stretches' the purple 'excess thrust' vector. Steepest climb is thus achieved when the excess thrust equals the max excess thrust available.

Glossary
• Resultant force: also called net force. The sum of all forces that act on the aircraft.
• Steady state flight: all acceleration (in Earth's reference frame) is zero. I.e., level flight, steady climb, steady descent. Due to $$\vec{F}=m\vec{a}$$, zero acceleration means zero resultant force.

Addendum I: From the comments, "A Piper Cub cannot take part of its thrust and apply it for lift." We see in the image above that sadly, that is exactly what a Piper Cub must be doing to climb at all. The excess thrust $$T_e$$ requirement $$T_e = W \sin \beta$$ with $$W$$ lift and $$\beta$$ the climb angle. If very little excess thrust is available, we had better keep $$\beta$$ small - exactly what you would expect for a little cub!

Addendum II: below, the same image but with excess thrust according to definition 1. The vectors are no longer a 'closed triangle', and the net force is not zero. The aircraft is now accelerating.

Images created using IPE.

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– Federico
Nov 13 '20 at 20:21

In a steady-state climb, Thrust is not a leg of the closed triangle of force vectors; rather, (Thrust-Drag) is.

See the right-hand vector diagram in this related answer-- Is excess lift or excess power needed for a climb?. The diagram shows that if our definition of "excess thrust" is (Thrust minus Drag), then excess Thrust clearly does exist in a steady-state climb.

The present question seems to imply that if we are flying horizontally, and then we increase Thrust, we must increase angle-of-attack if we wish to climb rather than accelerate horizontally. This isn't actually the case, at least over the long run. If we hold angle-of-attack constant and increase Thrust, we'll have a brief initial acceleration, but we'll end up in a steady-state climb at a slightly lower airspeed than we started with, at least in the case where little or no downthrust is present in relation to the actual direction of the flight path at any instant.

The reason for this is that for any given angle-of-attack, the lift and drag coefficients are constant, so airspeed scales according to the square root of the Lift vector, which also means that airspeed scales according to the square root of the Drag vector. And we know that Lift is slightly less in a steady-state climb than in wings-level flight, according to the relationship Lift = Weight * cosine (climb angle), at least in the case where no downthrust or upthrust is present relative to the direction of the flight path.

Note the relationship between Thrust and airspeed on either of the two right-hand tables in this related answer-- Is excess lift or excess power needed for a climb?.

The right-hand tables assume that no upthrust or downthrust is present, relative to the direction of the flight path. Each table assumes a constant angle-of-attack from one row to the next.

(But the airspeed values are not intended to be scaled in the same units between the two tables, since the lift coefficient is not specified, and is presumably much less in the upper table than in the lower table.)

You may find the vector diagrams, calculations, and explanations in this other related answer to be helpful as well -- Does lift equal weight in a climb?

When less than full thrust is in use during any steady state phase of flight, it can be considered that excess thrust is available.

Application of some or all of this excess thrust will result in a disturbance to the steady state - either acceleration in level flight, transition to a climb, increased climb rate, reduced descent rate or a combination, depending on what the pilot wants the airplane to do, eg at 10,000ft AMSL or FL100 accelerate from 250 kts and climb, once at climb speed, say 290 kts maintain the spd and continue climb at the resulting climb rate based on a fixed "climb thrust" setting throughout this sequence..

Just to clarify, there can be a steady state even during climb and descent if at constant spd and constant rate of climb and constant thrust, though in truth, this equation is also affected by the changes in properties described by the standard atmosphere.

Another way to look at this is that the calorific value of the fuel burnt by increasing thrust, puts energy into this system which is used for an increase in either kinetic energy (speed), or potential energy (altitude) or is shared by both.

At or close to the aircraft ceiling limitation, other factors come into play as also much excess thrust may just not be available, so that's another story.

• Re :When less than full thrust is in use during any steady state phase of flight, it can be considered that excess thrust is available."-- isn't it pretty clear that that's not the sense of "excess thrust" intended by the original question? Nov 12 '20 at 22:56
• Calorific value of the fuel burnt ... puts energy into the system, which is used for an increase in either (speed) or (altitude), yet they continue to show the thrust vector in their diagrams. Its the effect of thrust (pouring energy in) and how the increased drag is used, increased speed or altitude (increased AOA). Doesn't sound like you get less lift out of doing either one of those. Nov 12 '20 at 23:55
• @skipper44 I liked your practical answer and the fact that you mentioned changes in the properties of the atmosphere. I "fly my airspeed, not my throttle" because the planes aerodynamic qualities are dependent on speed. It just concerns me a bit that folks are talking thrust, when excess thrust input may not be fast enough to increase V if the plane is stalling low and slow. Excess thrust is converted into speed by F=ma. Nov 16 '20 at 8:48
• @RobertDiGiovanni, Thrust comes from the engine, which in turn runs on fuel. To increase thrust you have to increase fuel-flow. Achieving and maintaining a higher speed or altitude is entirely dependent upon the "calorific value of the fuel burnt by increasing thrust . . . " Nov 16 '20 at 8:52
• That's true, I could also spill aviation fuel on the ground and light it, but that won't make the plane fly (lots of calorific value though). But now we're talking engine efficiency. Thrust can only accelerate the aircraft. So, if one is too slow, and can't accelerate fast enough, they are $Screwed$. Nov 16 '20 at 8:57