justanotherflyer

28th Oct 2004, 22:33

Can anyone post (or post a link to) a simple, intuitively obvious way of explaining to beginning level students, why Vx is a function of excess thrust, and Vy of excess power.

View Full Version : Vx / Vy and all that

justanotherflyer

28th Oct 2004, 22:33

Can anyone post (or post a link to) a simple, intuitively obvious way of explaining to beginning level students, why Vx is a function of excess thrust, and Vy of excess power.

Keith.Williams.

29th Oct 2004, 08:39

Most aerodynamics books include the appropriate Drag/Thrust and Power available/required curves, but for many people these are far from intuitively obvious.

VX is the speed at which achievable angle of climb is greatest. We can get a mental picture of why climb angle is related to thrust and drag, if we consider two extreme cases

In straight and level flight the lift acts straight upwards. So all of the weight is carried by the lift. This means that the thrust required is equal to the drag.

In a vertical climb any lift would act horizontally, so all of the weight must be carried by the thrust. But we still have drag, so the thrust required is equal to the drag plus the weight.

As climb angle increases between zero (straight and level) and 90 degrees (vertical), the proportion of the weight that must be carreid by the thrust gradually increases from 0% to 100%.

The proportion of the weight that is carried by the thrust is equal to the weight times the sine of the climb angle. So at any climb angle, (including zero) the required thrust is equal to the drag plus the weight times the sine of the climb angle.

At zero degrees climb angle, the sine is zero so thrust required = drag.

The sine of 90 degrees is 1, so at 90 degrees climb angle the thrust required = drag plus the weight.

This means that the climb angle that we can acheived is proportional to the amount by which thrust exceeds drag. So our best climb angle is achieved at the speed where thrust exceeds drag by the greatest margin. To find this speed we need simply plot thrust and drag against speed, then find the speed at which the excess thrust is greatest.

VY is the speed at which rate of climb is greatest. To understand why this is related to power we need to start with some basic definitions.

Work is done when a force moves its point of application in the direction of the force. The amount of work done is equal to the force applied times the distance moved.

When an aircraft is in straight and level flight, the thrust must equal the drag. So the work done is equal to the thrust (or drag) times the distance flown through the air.

Power is the rate of doing work, which means that it is equal to the work done divided by the time taken to do that work.

So for our aircraft in straight and level flight, the power being used is equal to the thrust x distance flown/time taken. But distance/time = TAS, so power being used = thrust times TAS.

The power that is available for straight and level flight at any given TAS is equal to the thrust times TAS. And the power that is required is equal to the drag times the TAS. Provided thrust is equal to drag in straight and level flight at any given TAS, then we have sufficient power available to fly at that TAS.

If we now want to add a vertical speed component (ROC) we must provide some additional power to push the aircraft upwards. The force required to push the aircraft vertically upwards = weight and the vertical speed is the ROC. So the additional power required = weight times ROC.

This means that our best achievable ROC is proportional to the amount by which our power available exceeds our power required for straight and level flight. The greater the excess power available, the greater will be our best ROC. VY is the speed at which this excess power is greatest, so flight at VY gives best achievable ROC.

VX is the speed at which achievable angle of climb is greatest. We can get a mental picture of why climb angle is related to thrust and drag, if we consider two extreme cases

In straight and level flight the lift acts straight upwards. So all of the weight is carried by the lift. This means that the thrust required is equal to the drag.

In a vertical climb any lift would act horizontally, so all of the weight must be carried by the thrust. But we still have drag, so the thrust required is equal to the drag plus the weight.

As climb angle increases between zero (straight and level) and 90 degrees (vertical), the proportion of the weight that must be carreid by the thrust gradually increases from 0% to 100%.

The proportion of the weight that is carried by the thrust is equal to the weight times the sine of the climb angle. So at any climb angle, (including zero) the required thrust is equal to the drag plus the weight times the sine of the climb angle.

At zero degrees climb angle, the sine is zero so thrust required = drag.

The sine of 90 degrees is 1, so at 90 degrees climb angle the thrust required = drag plus the weight.

This means that the climb angle that we can acheived is proportional to the amount by which thrust exceeds drag. So our best climb angle is achieved at the speed where thrust exceeds drag by the greatest margin. To find this speed we need simply plot thrust and drag against speed, then find the speed at which the excess thrust is greatest.

VY is the speed at which rate of climb is greatest. To understand why this is related to power we need to start with some basic definitions.

Work is done when a force moves its point of application in the direction of the force. The amount of work done is equal to the force applied times the distance moved.

When an aircraft is in straight and level flight, the thrust must equal the drag. So the work done is equal to the thrust (or drag) times the distance flown through the air.

Power is the rate of doing work, which means that it is equal to the work done divided by the time taken to do that work.

So for our aircraft in straight and level flight, the power being used is equal to the thrust x distance flown/time taken. But distance/time = TAS, so power being used = thrust times TAS.

The power that is available for straight and level flight at any given TAS is equal to the thrust times TAS. And the power that is required is equal to the drag times the TAS. Provided thrust is equal to drag in straight and level flight at any given TAS, then we have sufficient power available to fly at that TAS.

If we now want to add a vertical speed component (ROC) we must provide some additional power to push the aircraft upwards. The force required to push the aircraft vertically upwards = weight and the vertical speed is the ROC. So the additional power required = weight times ROC.

This means that our best achievable ROC is proportional to the amount by which our power available exceeds our power required for straight and level flight. The greater the excess power available, the greater will be our best ROC. VY is the speed at which this excess power is greatest, so flight at VY gives best achievable ROC.

justanotherflyer

31st Oct 2004, 18:12

Thanks for your very detailed and helpful exposition.

I meant to post this to the instructors' forum - will request moderator to move it - the floor is open to any other instructors to relate how they teach these concepts.

I meant to post this to the instructors' forum - will request moderator to move it - the floor is open to any other instructors to relate how they teach these concepts.

QNH1013

31st Oct 2004, 19:11

Keith, Thanks for taking the time to post a detailed and helpful reply Really helped me understand better too.

Captain Stable

31st Oct 2004, 22:16

This is too good a topic to let the destructors have it all to themselves, so I won't move it. I'll put a pointer in Instructors' Forum to let them know it's here.

ROB-x38

2nd Nov 2004, 05:52

Another way to explain why Vx is a function of excess thrust is simply to look at what happens to the component of weight opposing the direction of flight in a climb (ie: resolve weight into a component opposite to lift and a component opposite thrust / with drag).

As the angle of climb is increased the component of weight opposing the direction of flight increases. As a result thrust must be increased.

This is quite simple but clearly shows that there is a direct relationship between thrust and the angle of climb.

The more thrust you have in reserve (excess thrust), the steeper you can make the angle.

--------

Great explanations Keith. I've got a question(s) you might be able to answer re: the actual speed for Vx -

Based on the fact that the thrust produced from a prop is maximum at low speeds, AFAIK the speed Vx is set at the lowest possible 'safe speed' to take advantage of this maximum excess.

So I was wondering how the speed for Vx is actually set in the flight manual?

Is it simply determined through flight testing?

Is there any requirement that it maintain a certain buffer above the power on/off stall speed?

Would it be true to say that the best AoC could be achieved right on the stall speed or would the maximum excess more likely be found at a speed just a bit below minimum drag (where drag has increased only slightly but the thrust has increased further at the lower airspeed)?

Cheers :ok:

As the angle of climb is increased the component of weight opposing the direction of flight increases. As a result thrust must be increased.

This is quite simple but clearly shows that there is a direct relationship between thrust and the angle of climb.

The more thrust you have in reserve (excess thrust), the steeper you can make the angle.

--------

Great explanations Keith. I've got a question(s) you might be able to answer re: the actual speed for Vx -

Based on the fact that the thrust produced from a prop is maximum at low speeds, AFAIK the speed Vx is set at the lowest possible 'safe speed' to take advantage of this maximum excess.

So I was wondering how the speed for Vx is actually set in the flight manual?

Is it simply determined through flight testing?

Is there any requirement that it maintain a certain buffer above the power on/off stall speed?

Would it be true to say that the best AoC could be achieved right on the stall speed or would the maximum excess more likely be found at a speed just a bit below minimum drag (where drag has increased only slightly but the thrust has increased further at the lower airspeed)?

Cheers :ok:

Keith.Williams.

2nd Nov 2004, 19:57

ROB

I have never been involved in the testing of prototype aircraft nor in the writing of flight manuals, so I cannot say with any certainty how it is done. But almost everything in aerodynamics is a matter of making a prediction then proving (or disproving) it by test flying.

In the manuals that I have examined these speeds have been stated as single valuse, or in tables to allow for things like varying altitude and temperature.

Best angle of climb at the stall is unlikely to be the best achievable overall. You are correct in saying that both thrust and drag are increasing as we decelerate close to the stall. But if you look at a typical pair of curves, the thrust curve is comparatively shallow in that area, while the drag is curve is very steep.

I have never been involved in the testing of prototype aircraft nor in the writing of flight manuals, so I cannot say with any certainty how it is done. But almost everything in aerodynamics is a matter of making a prediction then proving (or disproving) it by test flying.

In the manuals that I have examined these speeds have been stated as single valuse, or in tables to allow for things like varying altitude and temperature.

Best angle of climb at the stall is unlikely to be the best achievable overall. You are correct in saying that both thrust and drag are increasing as we decelerate close to the stall. But if you look at a typical pair of curves, the thrust curve is comparatively shallow in that area, while the drag is curve is very steep.

ROB-x38

2nd Nov 2004, 22:09

OK thanks Keith.

So most likely it's a compromise somewhere on the back of the drag curve which is then confirmed in flght testing.

So most likely it's a compromise somewhere on the back of the drag curve which is then confirmed in flght testing.

hawk37

3rd Nov 2004, 20:27

Rob, with a propellor aircraft, there can be a significant lift increase due to the slipstream from the propellor over the wings, compared to a "thrust only" powerplant such as a jet engine. Thus actual Vx for a propellor aircraft (at maximum power) will be less that the Vx one would calculate from the thrust/drag curves.

Hawk

Hawk

pa42

5th Nov 2004, 14:33

How interesting! In 50 years of flying FAA-certified aircraft using FAA-generated aeronautical theory texts, up to Aerodynamics for Naval Aviators (US Navy), I had never been exposed to thrust as an ingredient to Vx!

I believe (as a former FAA Designated Examiner) that in this corner of the globe, THRUST is not mentioned! All explanations of Vx are based on the POWER-required curve, Vx being found at the minimum-power-required (top of polar curve) speed, Vy being found at the point of tangency of a line from origin to polar curve. Similarly in Gliders (minimum sink vs best L/D).

Add this conundrum to the fact that French helicopter rotors revolve in the opposite direction to USA helicopter rotors. Vive la difference!

Dave

ATP CFI ASMELIGH

I believe (as a former FAA Designated Examiner) that in this corner of the globe, THRUST is not mentioned! All explanations of Vx are based on the POWER-required curve, Vx being found at the minimum-power-required (top of polar curve) speed, Vy being found at the point of tangency of a line from origin to polar curve. Similarly in Gliders (minimum sink vs best L/D).

Add this conundrum to the fact that French helicopter rotors revolve in the opposite direction to USA helicopter rotors. Vive la difference!

Dave

ATP CFI ASMELIGH