Maximum distance the airplane can cover

[aahil123]

Summary:: Maximum distance plane can cover

A General Aviation plane of 3,994 kg flies at its best lift/drag ratio with constant speed at an altitude of 3,817 m. The engine provides a thrust of 1,493 N to overcome drag.

Suddenly the engine fails. What is the maximum distance the plane can cover to find a suitable emergency landing spot (in km)?

I'm very confused and cannot find any equation relating this situation. Can i use SUVAT or any tips please? thanks

 

[anorlunda]

You turn the plane into a glider. From the PF Insights article:

But a typical Cessna 172 has a glide ratio of 1000 feet per 1.5 nautical miles which means a ratio of about 9:1. 9 feet horizonal for 1 foot vertical. An Airbus 320 has a glide ratio ~15:1. Modern racing gliders have a ratio ~50:1. Starting from 2 miles up, a glider could glide for 100 miles before reaching the ground.

Source https://www.physicsforums.com/insights/introduction-to-pitot-static-systems-in-gliders/

So if an Airbus 320 started at an altitude of 3,817 m, in theory it should be able to glide 15* 3817 = 57255 m before reaching the ground.

For a real life example of this scenario see https://en.wikipedia.org/wiki/Air_Transat_Flight_236
That flight lost engines at 12,000 m altitude and it glided 120 km to the Azores airport.

 

[aahil123]

where did you get 15 from?

 

[hutchphd]

The lift to drag ratio turns out to equal the glide ratio (you need to think about that for a while...in order to overcome drag you pay with altitude). So what is the lift to drag ratio here? be careful of the units!

 

[aahil123]

Is it not 10:1 for wvery 10 miles it travels forward it looses 1m in altitude?

 

[hutchphd]

Lift = weight of airplane.
Drag is given (engine overcomes drag when running)
Calculate ratio

 

[anorlunda]

Is it not 10:1 for wvery 10 miles it travels forward it looses 1m in altitude?

No. 10 meters forward for 1 meter loss in altitude. Or 10 miles forward for 1 mile loss in altitude.

The same distance units for forward motion as vertical motion.

The ratio 10 is only approximate. It depends on very many factors and it is measured, not calculated.

 

[willem2]

No. 10 meters forward for 1 meter loss in altitude. Or 10 miles forward for 1 mile loss in altitude.

The same distance units for forward motion as vertical motion.

The ratio 10 is only approximate. It depends on very many factors and it is measured, not calculated.

Well it can be calculated from the information given in the problem, so there's no need to assume any glide slope.
You should think about where the energy to overcome friction has to come from, when there is no engine.

 

[anorlunda]

Well it can be calculated from the information given in the problem, so there's no need to assume any glide slope.

The engine power prior to failure plays no role unless we add the restriction that airspeed remains constant after engine failure, but such a constraint makes no sense at all.

Any glider (including an airliner with failed engines) has a characteristic curve like this.

The plot shows horizontal speed versus vertical speed (altitude loss). The dotted lines in this example show 51.2:1 as the best ratio to go as far as possible (which is the OP question).

Part of the job of the pilot is to adjust airspeed appropriately to reach the target landing spot at the target's altitude.

 

[willem2]

The engine power prior to failure plays no role unless we add the restriction that airspeed remains constant after engine failure, but such a constraint makes no sense at all.

But there is such a restriction. The problem states that the plane already flies at the best lift/drag ratio, so the speed should remain the same. The rest of your post isn't useful, because we don't know what the characteristic curve is.

 

[hutchphd]

The engine power prior to failure plays no role unless we add the restriction that airspeed remains constant after engine failure, but such a constraint makes no sense at all.

I think you know too much about this stuff.

I believe the gist of the problem is to assume the engine power in level flight allows an estimate of the lift /drag ratio.
The best possible glide ratio (in my simplified understanding) is that same number.
Then it is just arithmetic. I'm certain this is not an accurate treatment...but if your engine is out you can't be choosy!

 

[anorlunda]

Maybe I do know too much. The best L/D with the propeller blowing air over the wings and the best L/D without the propeller seem like two different things to me. But you guys want to assume they are the same.

Is this supposed to be a homework problem? I thought the OP was just making stuff up, and therefore not a "solvable" problem statement.

 

[hutchphd]

In my happily simplified world the engine just pulls the plane forward without other effects.
Is my further supposition that the L/D gives a measure of the optimum glide ratio sensible? Then the information presented gives a simple answer (that seems a reasonable number to me)

 

[anorlunda]

I think we need the OP to clarify the nature of the problem. Is it homework, or is it a bar bet, or is it curiosity?