# How can I calculate the maximum range of an aircraft descending with power (not gliding)?

Let's assume I have an aircraft, for which the thrust is not enough to keep the altitude. So over time I will lose some altitude. But it will still go further then it would without having an engine (gliding).

How can I calculate the maximum possible range for this case?

I know the Breguet equation, but I do not think this helps me in this case?

Some formulas would be great.

• I think you may find interesting elements on the section about power curve of the resource "how it flies?" – Manu H Feb 5 at 8:17

You can use the lift to drag ratio (L/D) to calculate for just about any stage of flight if you compensate it properly. Assuming your descent is steady-state, lift is equal to weight, using the L/D you can then determine drag. Next take drag and subtract the thrust produced by the engine to get the amount of excess drag need to be covered by gravity. You glide ratio will be lift divided by excess drag. This makes for a simple calculation of your variables remain constant, however you will need to use an integral if you want to compensate for changes such as differences in engine power at different altitudes, and reduction in weight as fuel is burned.

You ultimately want the glide angle, so we can just look at the equilibrium equations for the glider as a starting point. I will assume a quasi-stationary, wings-level.

$$L=W\cdot \sin{\gamma}$$ $$D=W\cdot \cos{\gamma}$$

where $$\gamma$$ is the glide angle, positive downwards. This leads to the self-evident: $$\tan{\gamma}=\frac{1}{E}$$ Which just says that the glide angle becomes shallower with increasing aerodynamic efficiency. If we add in a thrust component to our glider, we can redo the above for a powered glide:

$$L=W\cdot \sin{\gamma_p}$$ $$D=W\cdot \cos{\gamma_p}+T$$

which gives: $$\tan{\gamma_p}=\frac{D-T}{L}= \frac{1}{E} - \frac{T}{L}$$ This is dimensionally sound and, more importantly, converges to both the original glide angle equation for $$T=0$$ and to $$\gamma=0$$ (horizontal flight) for $$T=D$$. For either angle, knowing your starting altitude, you can obtain the glide range as: $$x_{glide}=\frac{h}{\tan{\gamma}} \left( \approx \frac{h}{\gamma} \space \text{for small angles} \right)$$

• Well, sir, at steady state, (no acceleration, forces balanced), (using the Lift equation), does $Lift = Weight$ at a given airspeed and trim regardless of orientation. – Robert DiGiovanni Feb 8 at 0:02
• Lift does not equal weight in a glider, you got a number of decent answers on that front here, maybe you should review them? – AEhere supports Monica Feb 8 at 0:27
• well, as we advance our state of the art, we come to realize Lift has NO relationship to climbing, descending, powered or gliding. It is based on coefficient of lift (airfoil type and AOA), airspeed, and wing area. So, former answers go by the wayside, or are improved. Continue to learn so much here. I will review your equations. – Robert DiGiovanni Feb 8 at 5:03

At a given airspeed, your plane (not only the wings, but the entire plane) will have an optimal L/D ratio. That's also the glide ratio. For example, if the best L/D is 10, your airplane, flying in a glide, will move horizontally (at the airspeed for optimal L/D), 10 kilometers for every kilometer of altitude that it loses.

• Since engine thrust will improve with increasing density, the effective L/D will also improve as the aircraft descends. – Peter Kämpf Feb 4 at 20:03
• Okay but an aircraft, flying at best L/D will reach a bigger distance with a switched on engine than with a switched off engine. – Lucas Feb 4 at 20:09
• How do I account for this? And besided: a commerical aircraft, is it always flying with best L/D during horizontal flight? – Lucas Feb 4 at 20:11
• This answer would be correct for a glide, but power confounds the L/D condition. – Zeiss Ikon Feb 5 at 15:11