VMP VMD contradiction? Please help!

[baleares]

I am having trouble trying to rationalise the following.

If you fly at any speed below VMD you will need to use more thrust according to the total drag curve. [edit for clarity: I am talking about the FORCES curves of THRUST REQ./DRAG)

But, the speed for max. endurance (least fuel use and time in the air), which I understand to be VMP, is below VMD.

How can you have to use more thrust yet still use less fuel ?

[Mach Jump]

How can you have to use more thrust yet still use less fuel ?

You can't.

Unless you have a very strange engine that has an increasing fuel flow with reducing thrust, VMD=Max End Speed. (Assuming we're working with the same kind of speed in both instances, ie. IAS, CAS, etc.)

I think confusion arises here when things are quoted in different kinds of speed, eg. power curve in TAS and drag curve in EAS.


MJ

[Level Attitude]

I think you are mixing up your Minimums and your Maximums; and your
Drag and your Distances.

For maximum time in air for a given amount of fuel.
V Min Power = V Min Drag = V Max Endurance = The speed corresponding
to the lowest point on a Speed vs Drag Curve.

For maximum distance flown for a given amount of fuel.
V Max Distance is always slightly higher than V Min Drag and is the speed
corresponding to where the Tangent touches the Speed vs Drag Curve.

Sorry, do not know how to attach images (of a V/D curve) which would
easily illustrate this.

[baleares]

Just to clarify, I am referring to a prop driven piston engine...

I have been taught that Endurance is VMP (bottom of power required curve) and Range is VMD (tangent of power required curve). Never heard of V max distance!!!

[Mach Jump]

Here are a couple of posts from the archive on the subject. see what you make of them.

no sponsor31st Oct 2004, 14:41
It is a very complex subject. Work this out, and you will nail the Perf exam, and a good bit of PoF.

VMD - Minimum Drag. Found on the drag curve, at the bottom. Best speed for Endurance (Jet) Minimum glide angle, and it is also where CL:CD is at a maximum, or CD:CL is at a minimum. Look on the drag curve and you'll see it is at the point where induced and profile drag cross. It is also equal to a bunch of other things, such as Vx for a Jet, and V4 (v2+10).

VMP is min power. So you need to look on the power curve. It is at the bottom on the power req curve. Here you want to use as little fuel as possible, so for the prop it will be almost near Vs, and very much in the speed unstable regime. This is where gliding for endurance for Prop and Jet will be. If you want to find VMD, it will be at a tangent on the pwr req curve. So you can conclude that VMP is slower than VMD.

Best speed for range will be at the tanget of a line from the origin which just touches the drag curve. It will be at 1.32 Vmd. Most jets will fly a little higher than this at 1.37 Vmd, since it equates to 99% range, but an increase in 4% speed, which makes it better economically.

What you will need to know for the exam is how these things change. What happens to VMD as a EAS and as a TAS at altitude?

Beware that the Power curve is in TAS and drag in EAS, so all EAS speeds don't change, but TAS speeds do as you get higher. So, you need to learn the inter-relationships between EAS, RAS, TAS and Mach numbers. Remember ERTM - if EAS is constant and you climb, then RAS, TAS and Mach will all increase.

If you can figure out profile and induced drag, and how these are affected by flaps, increases in mass etc, then you will be well on your way.

As I said, it's complex. Try the book, 'Performance' by Swatton.

--------------------------------------------------------------------------

Keith.Williams.31st Oct 2004, 15:03
VMP is the speed at which the power required in minimum

VMP is also the speed at which propeller aircraft endurance is greatest. This is because fuel flow in propeller aircraft is proportional to power. So by flying at the minimum power required speed we use the minimum fuel flow. Using minimum fuel flow maximises the time it takes to burn the fuel on board.

VMP is also the speed at which the glide endurance is greatest. This is because in flight the aircraft expends energy moving forward against the drag force. The total energy of the aircraft is the sum of its kinetic energy (1/2m Vsquared) + potential energy (weight x height). In powered flight this energy is constantly being replenished by the engine. But if we shut down the engine and glide, the stored energy cannot be replenished. So as we glide we gradually use up the stored energy. When we are stoped on the ground we have used up all of the stored energy. So for maximum glide endurance we must use up the stored energy as slowly as possible. Energy useage rate is power required. So for maximum glide endurance we must fly at the minimum power required speed (VMP).

VMD is the speed at which total drag is least. It is also the speed at which we achieve:

a. Best propeller aircraft range.
b. Best glide range.
c. Best jet aircraft endurance.

We can find Vmp from a power required curve by drawing a tangent from the origin to the curve. This will touch the curve at Vmd. We can find Vmp from a drag curve by drawing a straight line from the origin to hit the curve at 90 degrees. The line will cross the curve at Vmp. If you do this a few times you will find that Vmp is always less than Vmd.

If you want an easy way to remember how to find the speeds for best range and best endurance for props and jets just remember the terms:

"P for Power P for Props" and "D for Drag D for Dyets".

For prop aircraft "P for Power P for Props" means that you sketch the curve for power required against airspeed. It looks a bit like a Nike tick. If you draw a horizontal line to just touch the bottom of the curve, it will touch it at Vmp. This is the speed for best prop endurance. Now draw a straight line from the origin to just touch the curve. It will touch it at Vmd which is prop best range speed. The horizontal line gives best endurance and the tangent give best range.

For a Jet just do the same but this time use the drag curve. (D for drag D Dyets). The horizontal line will touch the bottom of the drag curve at Vmd which give best jet endurance. The tangent drawn from the origin touches the curve at about 1.32 Vmd which is best Jet range speed. Once again the horizontal line gives best endurance and the tangent gives best range.

For the ATPL exams it is also worth noting that best glide endurnce means lowest rate of descent. They sometimes use the terms "Vmsr, Velocity minimum sink rate" or "Vmdr , Velocity Minimum descent Rate" insted of Vmp. When gliding at maximum range speed your glide angle will be minimum so they sometimes use the term "Vmga, Velocity Minimum Glide Angle" instead of Vmd.

I'm with you guys. I can't quite see why the Power Required, Thrust Required, and Total Drag curves are not the same, assuming we are using the same speed, ie. IAS, TAS, CAS, etc. and, perhaps because I cant see that, I dont understand how VMD can be the speed for Max Range. Maybe someone smarter than us will come up with an explanation that us simple folk can grasp!


MJ

[Level Attitude]

Never heard of V max distance!!!

Range = How Far = Distance

Having said that I gave a very quick, oversimplified response.

I tend to think of Drag = Thrust = Power, and the last is not correct.

I do have a diagram to illustrate, but I have no idea how to insert it.
So am now excusing myself from this thread.

[Mach Jump]

Right, got it now.

The drag, and thrust required curves are the same, but the power required curve is not. (see above diagram.) The power required curve also represents the fuel flow required, and as the min. fuel flow is the best endurance speed,(100kt TAS) then it has to be lower than min. drag speed (130kt TAS).

Where a line drawn from the origin of the graph makes a tangent to the power/fuel required curve is the best range speed. as can be seen, this corresponds with the min. drag speed.


MJ

[baleares]

Thank you all so much for your input.

Looking at the THRUST / DRAG curve (which are curves of force, not power). The speed for the use of least THRUST is the speed of MIN DRAG. To fly any faster or slower than V MIN DRAG you will have to use more thrust.

But thrust is fuel!

According to my books and what I have been taught V MIN DRAG is found on the power curve by drawing a straight line from the origin until it makes a tangent with the POWER REQUIRED curve.

This makes V MIN DRAG faster than V MIN POWER.

But we know from the THRUST/DRAG curves that to fly at any other speed than V MIN DRAG requires more THRUST.

How can more THRUST mean less POWER and therefore less fuel?

[Level Attitude]

But thrust is fuel!

NO. As you quite rightly have stated previously Thrust is a Force.

A propeller will consume Power to produce this Thrust, but this
Power comes from that produced by the engine.

Fuel use depends on Engine Power but, due to inefficiencies, this will not
be the same as the Power available to the propeller.

How can more THRUST mean less POWER and therefore less fuel?

Unsure if this is the complete and correct answer but:
Propeller efficiency (how much engine Power it can use to turn in to Thrust)
varies with speed - so it is possible for a Propeller to produce more Thrust for
less Engine Power and hence less fuel consumption.

[Tinstaafl]

Bear in mind that for a piston engine, fuel flow is approximately proportional to *power* output. That power is then converted to thrust by a propeller - and speed affects its efficiency. For a jet, FF is approximately proportional to *thrust* output.

Make sure you're comparing the correct curves for the engine the aircraft has.



<Level Attitude was faster to write & more eloquent than I>

[BEagle]

For range, fuel burn v. distance is required to be a min. value, whereas for endurance it's fuel burn v. time.

Thus for a jet aircraft, flight for range occurs at the max TAS/Drag speed, normally 1.32 Vmd.

1.32 is the fourth root of 3, incidentally. If you understand basic calculus, that is simply derived from comparing minimum values for Drag and Power, using the relationships D = A/VČ + BVČ and P=D x V, having differentiated D with respect to V and equating to zero, then the same thing for dP/dV and comparing the 2 minima.

[Mach Jump]

Maybe someone smarter than us will come up with an explanation that us simple folk can grasp!

1.32 is the fourth root of 3, incidentally. If you understand basic calculus, that is simply derived from comparing minimum values for Drag and Power, using the relationships D = A/VČ + BVČ and P=D x V, having differentiated D with respect to V and equating to zero, then the same thing for dP/dV and comparing the 2 minima.

Beagle.

My days of differential calculus are very long ago, and little used since.

I'm reminded of a story told by (or about) Former Master of the Rolls, Lord Denning.

The Judge had asked a question of an expert witness, and having endured a long and technical explanation, looked despairingly at Council and said, 'Frankly Mr Xxxxxx, I'm none the wiser.'

Council, without hesitation, stood and replied, ' No Mi'lord, but you are now much better informed!'


MJ

[keith williams]

How can more THRUST mean less POWER and therefore less fuel?

To examine this let’s go back to basics.

Hopefully we can all agree that:

Power required Drag x TAS

And

For straight and level constant speed flight Thrust = Drag

This means that we can use the Drag / TAS curve to test how power required varies with drag.

Now let’s look at the curves in post number 7.

To keep the number small We will ignore the numbers of the scales and use the numbers of squares to measure values of drag, TAS and power.

At 2 squares along the TAS axis we have 17 squares up the drag axis. So the power required at this speed is 2 x 17 = 34.

At 4 squares along the TAS axis we have 5 squares up the drag axis. So the power required at this speed is 4 x 5 = 20.

Continuing this process w get the following results:

TAS------- Drag --------Power Required
2---------- 17 ---------- 34
4----------- 5 ---------- 20
4.5 --------- 4 --------- 18.5
5 ----------- 3.7 ------- 18.5
7 ------------3.3 23.1
10 ---------- 4.3 ------- 43
15 ---------- 7.3 ------- 109.5

Note that the position of the power required curve in post 7 does not reflect these values. This is because it has been positioned to suit the power required scale at the left edge of the diagram.


Between 2 and 4 drag units the drag and power required both decreases. This is because the rate at which the drag is decreasing is greater than the rate at which the TAS is increasing. So the product of TAS x Drag is decreasing.

Between 4.5 and 5 drag units the drag is decreasing but the power required is constant. This is because the rate at which the drag is decreasing is equal to the rate at which the TAS is increasing. So the product of TAS x Drag is constant.

Between 10 and 15 drag units the drag and power required are both increasing. So the product of TAS x Drag is increasing.

Now to answer the specific question:

How can more THRUST mean less POWER?

Let's look at the effect of a deceleration from 7 squares on the TAS scale to 4 squares on the TAS scale.

At 7 TAS we have 3.3 drag so power required = 7 x 3.3 = 23.1
At 4 TAS we have 5 drag so power required = 4 x 5 = 20.
So although the drag (and hence thrust required) went up from 3.3 to 5, the power required went down from 23.1 to 20.

The key factor in determining whether power required increases, remains constant or decreases, is the relative magnitudes of the rates of change of TAS and Drag.

[BEagle]

Sorry, my fat fingers. Conventionally it's actually D=AVČ + B/VČ (where AVČ indicates the elements of drag which increase with speedČ and B/VČ indicates lift dependent drag which decreases with speedČ)

If D=AVČ + B/VČ

Then dD/dV = 2AV - 2B/Vł

At minimum, dD/dV = 0. Hence 2AV=2B/Vł, in other words Vmd=(4th root) B/A

But P=D x V; i.e. P=AVł + B/V

dP/dV = 3AVČ - B/VČ

Again, at minimum, dP/dV = 0. Hence 3AVČ=B/VČ, in other words Vmp=(4th root) B/3A.

Thus comparing these results, Vmd/Vmp = (4th root) 3 = 1.32

[keith williams]

For readers who do not like maths and like calculus even less, we can use the diagram in post number 7 to see why increasing drag does not necessarily cause an increase in power required, and also why Vmp is not the same as Vmd. (This method is just the poor man’s calculus)

Hopefully we can all agree that Power Required = Drag x TAS.

Looking at the diagram in post 7 we can see that TAS is increasing left to right along the horizontal axis and Drag is increasing from bottom to top on the vertical axis. The area within these two axes is marked off in squares.

Power required = Drag x TAS. But at any point on the Drag curve, Drag x TAS is also the area beneath the curve between the origin and that point on the drag curve. So we can calculate the power required at any point on the drag curve by counting the squares.

So, for example at Vmd (at the bottom of the Drag curve) we have an area made up of about 7 squares along the TAS axis and 3 squares up the Drag axis. This gives an area of 7 x 3 = 21 squares. This means that the power required at Vmd is proportional to an area of 21 squares.

Vmp is the speed at which the power required is the least. So Vmp must be the same speed as a point on the drag curve at which the area below the drag curve is the least.

If we count 5 squares along the TAS axis, we can see that there are about 3.6 squares up the Drag axis, so we have an area of 5 x 3.6 = 18 squares. This is less than the 21 squares at Vmd, which means that the power required at this new speed is less than at Vmd. So we have moved closer to Vmp.

If we count 4 squares along the TAS axis, we can see that there are about 5 squares up the Drag axis, so we have an area of 4 x 5 = 20 squares. This is more than the more than 18, so we have moved beyond Vmp. So Vmp is somewhere between 4 squares and 5 squares along the TAS axis.

If we count 4.75 squares along the TAS axis, we can see that there are about 3.75 squares up the Drag axis, so we have an area of 4.75 x 3.75 = 17.8 squares. This is lower than the previous value, so we have moved closer to Vmp.

We could keep on repeating this process until we reached a good approximation for Vmp. Or if we wanted to we could short-cut this process by drawing a straight line that started at the origin and crossed the drag curve at 90 degrees.

[BEagle]

I knew there was a good reason for the differential calculus I was taught at the age of 14....

Much easier than counting squares!

[keith williams]

If you think back to when you were 14 (to do this your memory will need to be lot better than mine) you will probably recall that the subject covered immediately before calculus was estimating areas under curves.

Calculus is after all simply a method for estimating areas under curves. Which is why I said in my previous post that, "This method (square counting) is just the poor man’s calculus".

More recently having spent 10 years teaching ATPL theory, I have found that the majority (yes I really do mean the majority) of modern-day ATPL students would stop listening at the very mention of calculus. But even the most mathematically challenged could count squares.

[Mach Jump]

Keith, Beagle

To most people, the suggestion that it takes less fuel to overcome more drag is, on the face of it, perverse.


I don't think it's about calculus, or counting squares. What people need is a practical explanation of what seems an illogical situation, and we should be able to think of a convincing way to give them one.


MJ

[keith williams]

To most people, the suggestion

that it takes less fuel to overcome more drag is, on the face of it, perverse.

The problem with the statement above is that is based on the assumption that the amount of fuel we burn must be determined by the amount of thrust we are producing. This is a false assumption.

I weigh about 210 lbs (yes I am a bit of a fat ……). If I sit or lie on the ground I exert a force of 210 lbs on the ground. Exerting this force uses up no energy. But if I were to repeatedly lift a 100 lbs weight above my head and lower it back to the ground, I would expend a great deal of energy. Isn’t that perverse?

So why does lifting the 100 lbs weight expend more energy than exerting a 210 lbs force on the ground? The amount of energy that my body expends in doing this type of thing is determined by the amount of mechanical work that I am doing. Exerting the 210 lbs force on the ground does not require me to do any mechanical work, so no energy is required (I don’t even need to stay awake). But repeatedly lifting the 100 lbs weight involves doing 100 ft lbs of work for every foot that I lift the weight above the ground.

The key fact in all of this is that the energy used is not determined by the force that I apply, but is determined by the amount of work that I am doing. Power is the rate of doing work, so the energy (fuel) consumption of my body is determined by the power required to do the work. Exactly the same situation applies to propeller aircraft. Fuel consumption is not proportional to the thrust produced, but is proportional to power output.


There are a great many aspects of aerodynamics and physics which appear to be perverse.

For example:

1. Fuel flow in a jet is proportional to thrust, but fuel flow in a piston/prop is proportional to power?
2. At speeds below Vmd, the faster we fly the less drag we have.
3. At speeds below Vmp, the faster we fly the less power we need.
4. For jet aircraft at speeds below Vmd, the faster we go the less fuel we burn.
5. For piston/propeller aircraft at speeds below Vmp, the faster we go the less fuel we burn.

Looked at individually each of these statements may appear to be perverse.


But if we rearrange the order of the statements slightly the situation becomes a little bit clearer.

6. Fuel flow in a jet is proportional to thrust.
7. At speeds below Vmd, the faster we fly the less drag we have.
8. For jet aircraft at speeds below Vmd, the faster we go the less fuel we burn.

Statement 8 is true because statements 6 and 7 are true. As we accelerate towards Vmd, the reducing drag requires us to produce less thrust, and because fuel flow is proportional to thrust, producing this reduced thrust reduces our fuel flow.


9. Fuel flow in a piston/prop is proportional to power.
10. At speeds below Vmp, the faster we go the less power we need.
11. For piston/propeller aircraft at speeds below Vmp, the faster we go the less fuel we burn.

Statement 11 is true because statements 9 and 10 are true. As we accelerate towards Vmp, the reducing power required enables us to produce less power, and because fuel flow is proportional to power, producing this reduced power reduces our fuel flow.

The real conundrums are why are statements 6, 7, 9 and 10 true.