ATPL : Performance
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ATPL : Performance
Hello all,
Here is a question from which i don't understand the answer.
On a twin engined piston aircraft with variable pitch propellers, for a given mass and altitude, the minimum drag speed is 125 kt and the holding speed (minimum fuel burn per hour) is 95 kt.
The best rate of climb speed will be obtained for a speed :
a) inferior to 95 kt
b) equal to 95 kt
c) is bteween 95 and 125 kt
d) equal to 125 kt
Good answer is B.
They say that the minimum drag speed Vmd is 125 kts. Now, we know that for propeller aeroplane Vmd is the best rate of climb speed (and the best range one as well). So logically in this question the right answer for the question "what is the best rate of climb speed" should be 125kt, isn't it? Why is it not the case? Could someone explain me why 95 kt (which is the holding speed) is the best rate of climb speed for this aeroplane?
Many thanks.
Here is a question from which i don't understand the answer.
On a twin engined piston aircraft with variable pitch propellers, for a given mass and altitude, the minimum drag speed is 125 kt and the holding speed (minimum fuel burn per hour) is 95 kt.
The best rate of climb speed will be obtained for a speed :
a) inferior to 95 kt
b) equal to 95 kt
c) is bteween 95 and 125 kt
d) equal to 125 kt
Good answer is B.
They say that the minimum drag speed Vmd is 125 kts. Now, we know that for propeller aeroplane Vmd is the best rate of climb speed (and the best range one as well). So logically in this question the right answer for the question "what is the best rate of climb speed" should be 125kt, isn't it? Why is it not the case? Could someone explain me why 95 kt (which is the holding speed) is the best rate of climb speed for this aeroplane?
Many thanks.
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Hi thibaut,
Quick one ~ been a while since I did Performance but the way I looked at the question it's in regard to Props not jets. Therefore you're going to be wanting your Vmp, minimum power speed. (which seems to be from the question your minimum fuel burn per hour)
Anyway I'm sure no doubt someone will be along shortly to give you a more comprehensive answer
Quick one ~ been a while since I did Performance but the way I looked at the question it's in regard to Props not jets. Therefore you're going to be wanting your Vmp, minimum power speed. (which seems to be from the question your minimum fuel burn per hour)
Anyway I'm sure no doubt someone will be along shortly to give you a more comprehensive answer
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The best rate of climb (Vy) for a prob is the biggest difference between power required and power available and is therefore generally between Vimp (min power speed) and Vimd (examples are usually but closer to Vimd). Therefore answer C is correct and B is wrong.
In that question, 95kts would be closer to Vx than Vy
In that question, 95kts would be closer to Vx than Vy
I'm confused the the best angle is is the greatest difference between power required and power available i.e highest ROC
and Vy is the greatest between Thrust Horse Power required and Thrust Horse Power available [slightly simplified]
where power is thrust * velocity so higher velocity means greater thrust horse power but and at Vimd you require the least thrust and least drag equating to the greatest spread of Pr and Pa,...in the question it appears that at the greater velocity [124 kts] therefore the greatest difference in THPr and THPa i.e Vy
maybe the question meant Vx
and Vy is the greatest between Thrust Horse Power required and Thrust Horse Power available [slightly simplified]
where power is thrust * velocity so higher velocity means greater thrust horse power but and at Vimd you require the least thrust and least drag equating to the greatest spread of Pr and Pa,...in the question it appears that at the greater velocity [124 kts] therefore the greatest difference in THPr and THPa i.e Vy
maybe the question meant Vx
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IMHO the examiner is trying to tell you something by specifying a powerful aircraft with VP propellers. For this aircraft In the lower speed ranges you would expect the THP available to be a nearly flat line, although I admit a constant speed prop would give a flatter trace. If you draw THP available as constant over this range and superimpose the power required curve you will see that the maximum power surplus occurs at minimum power speed, Vimp. You have been given this as 95kt
My whole argument relies on having THP available as roughly constant at these speeds, which is a special case. It would not necessarily apply in the real world or to a fixed pitch prop. The shape of the THP available curve (assuming constant EHP) depends on propeller efficiency versus TAS and on your ordinary bugsmasher rises rapidly from zero TAS and then falls off again
This speed is now your best rate of climb speed, Vy. Your best angle of climb speed, Vx, is where you have maximum excess thrust over drag and because props give maximum thrust at very low speed to find Vx you would fly as slowly as possible - minimum control speed is usually quoted
So I go for B
Dick
My whole argument relies on having THP available as roughly constant at these speeds, which is a special case. It would not necessarily apply in the real world or to a fixed pitch prop. The shape of the THP available curve (assuming constant EHP) depends on propeller efficiency versus TAS and on your ordinary bugsmasher rises rapidly from zero TAS and then falls off again
This speed is now your best rate of climb speed, Vy. Your best angle of climb speed, Vx, is where you have maximum excess thrust over drag and because props give maximum thrust at very low speed to find Vx you would fly as slowly as possible - minimum control speed is usually quoted
So I go for B
Dick
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For this aircraft In the lower speed ranges you would expect the THP available to be a nearly flat line, although I admit a constant speed prop would give a flatter trace.
THP = Thrust X TAS, so all aircraft propulsion systems produce zero THP when TAS is zero.
A constant speed prop will maintain thrust up to higher airspeeds, so the thrust line will be flatter.
But THP = Thrust X TAS, so a flat thrust line will give a THP line which increases with increasing TAS.
A constant speed prop will extend the THP curve further to the right.
This will increase the speed for best ROC.
For a very basic piston prop VX will be a smidgin less than Vmp and Vy will be a smidgin more than Vmp.
For higher powered aircraft, particularly those with constant speed props Vy will be closer to Vmd.
The correct only correct answer in this question is
c) is bteween 95 and 125 kt
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Sorry Keith, I am not confusing thrust and THP. On the other hand I think you have not understood my post, which offers for discussion a possible scenario. I don't think it helps to quote the standard section from your notes, which in Logic is called "appealing to revered authority" The examiner has something else in mind.
If the power available curve is flat at speeds near Vimp then Vimp gives the max excess of power. If the slope of the power available curve is rising with TAS then Max power excess will be higher than Vimp.
If the examiner gave 95kt as the correct answer then he presumably had a reason for it. Perhaps he has performance curves for a specific twin that he has used to make up the question. On the other hand it may just be a feedback error
How are things with you these days?
Dick
If the power available curve is flat at speeds near Vimp then Vimp gives the max excess of power. If the slope of the power available curve is rising with TAS then Max power excess will be higher than Vimp.
If the examiner gave 95kt as the correct answer then he presumably had a reason for it. Perhaps he has performance curves for a specific twin that he has used to make up the question. On the other hand it may just be a feedback error
How are things with you these days?
Dick
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I don't think it helps to quote the standard section from your notes, which in Logic is called "appealing to revered authority"
It is fact that constant speed props delay the fall off in thrust as speed increases. This creates a flatter thrust curve at the low speed end of the speed range.
It is also a fact that THP = Thrust X TAS.
So by producing a flatter thrust line, a constant speed prop causes THP to increase as speed increases (untill the eventual reduction in thrust outweighs the increasing TAS).
It is also a fact that because THP = Thrsut X TAS, all aircraft propulsion systems produce zero THP when TAS is zero.
If the power available curve is flat at speeds near Vimp then Vimp gives the max excess of power.
I agree, but that is a big IF.
If the slope of the power available curve is rising with TAS then Max power excess will be higher than Vimp.
If the examiner gave 95kt as the correct answer then he presumably had a reason for it.
Perhaps he has performance curves for a specific twin that he has used to make up the question.
We both know that many of the CQB questions have been and continue to be rubbish. This is just one more of those rubbish questions.
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Sorry again Keith, I don't mean to upset you. The big IF is the point of my post. Can anyone suggest a set of conditions that would give a flat power available curve at speeds near Vimp? Fairly obviously you think not, so that consigns the question to the bad question box.
Dick
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I'm not at all upset Dick.
It just appears to me that you have taken the question and attempted to justify the examiner's answer. I have looked at it, and considered what the correct answer is.
But let's look at your idea of a powerful engine and constant speed prop giving a flat THP curve at Vmp.
Most texts state that for a basic low powered aircraft with a fixed pitch prop, Vx is just less than Vmp and Vy is just more than Vmp.
For Vx we need maximum excess thrust. This means that as speed increases towards Vmp, the benefit of decreasing drag must outweigh the cost of decreasing thrust, such that excess thrust is maximum just below Vmp.
For Vy we need maximum excess power. So for Vy to be just greater than Vmp, this means that as speed increases to just beyond Vmp, the benefit of increasing THP available must outweigh the cost of increasing power required.
The important points here are that thrust is decreasing and THP available is increasing as we approach and pass Vmp.
To create a flat THP available curve at Vmp we would need to have a situation in which the thrust reduces at exactly the right rate to balance the increasing TAS, such that THP = Thrust X TAS = Constant.
Well, if we already have increasing THP at Vmp with a fixed pitch prop, why would we design a constant speed prop that causes the THP to be constant at Vmp?
I agree that a flat THP curve at Vmp would give Vx = Vmp. But I just cannot see why we would want to create a flat THP curve at Vmp, because to do so we would need to ensure that as speed increases, the thrust decreases even faster than it does with a fixed pitch prop.
I know that this type of question has been appealled in the past. If you cast your mind back to who was managing PERF for the CAA at that time, you will not be at all surprised that the "improved" question is still a mess.
It just appears to me that you have taken the question and attempted to justify the examiner's answer. I have looked at it, and considered what the correct answer is.
But let's look at your idea of a powerful engine and constant speed prop giving a flat THP curve at Vmp.
Most texts state that for a basic low powered aircraft with a fixed pitch prop, Vx is just less than Vmp and Vy is just more than Vmp.
For Vx we need maximum excess thrust. This means that as speed increases towards Vmp, the benefit of decreasing drag must outweigh the cost of decreasing thrust, such that excess thrust is maximum just below Vmp.
For Vy we need maximum excess power. So for Vy to be just greater than Vmp, this means that as speed increases to just beyond Vmp, the benefit of increasing THP available must outweigh the cost of increasing power required.
The important points here are that thrust is decreasing and THP available is increasing as we approach and pass Vmp.
To create a flat THP available curve at Vmp we would need to have a situation in which the thrust reduces at exactly the right rate to balance the increasing TAS, such that THP = Thrust X TAS = Constant.
Well, if we already have increasing THP at Vmp with a fixed pitch prop, why would we design a constant speed prop that causes the THP to be constant at Vmp?
I agree that a flat THP curve at Vmp would give Vx = Vmp. But I just cannot see why we would want to create a flat THP curve at Vmp, because to do so we would need to ensure that as speed increases, the thrust decreases even faster than it does with a fixed pitch prop.
I know that this type of question has been appealled in the past. If you cast your mind back to who was managing PERF for the CAA at that time, you will not be at all surprised that the "improved" question is still a mess.
hey fellas
but,....velocity is a vector and in the case of a climb you have two components the forward component along the flight path and the vertical component,..with no other data [certification parameters or propeller advance ratio Vmc etc,...it can only said that at Vy you want the highest forward component of velocity along with a vertical component that is based on the amount of excess THP available for climb,..in order to obtain the highest forward component you'd want the highest IAS [TAS] i.e the highest 'V' possible and the greatest excess of THP and that would occur at Vimd in which you have the lowest drag there for the lowest thrust required at the highest TAS possible giving the greatest difference between THPr and THPa,...
I think the question meant to say Vy not Vx as stated
old commentary on this:
I think the question meant to say Vy not Vx as stated
old commentary on this:
Because at that point in a piston [a constant power variable thrust machine--as Bookworm explained] you have the lowest drag coincident greatest possible lift coincident the only thing affecting climb rate as such is the difference between POWER required and POWER available, however as the power falls off with altitude while drag is always equal to thrust....
But, ROC will decrease so to maintain the highest possible ROC [the condition required of Vx] at less power [from critical altitude] an increase in the lift term and therefore angle of attack is required this action induces drag and requires greater thrust maintain that new AOA at a constant altitude,...i.e there's less actual drag at altitude, but more drag required because more lift is required, so more thrust is required,...READ CAREFULLY THIS!!!
And, since power is thrust X Vtas and increase in velocity is required to maintain the required power for max climb angle,...
so in pistons Vx continually increases until it converges with Vy at the absolute performance altitude, whereas a jet produces constant thrust and a constant the AOA is that which produces the greatest climb rate i.e the AOA for L/Dmax
But, ROC will decrease so to maintain the highest possible ROC [the condition required of Vx] at less power [from critical altitude] an increase in the lift term and therefore angle of attack is required this action induces drag and requires greater thrust maintain that new AOA at a constant altitude,...i.e there's less actual drag at altitude, but more drag required because more lift is required, so more thrust is required,...READ CAREFULLY THIS!!!
And, since power is thrust X Vtas and increase in velocity is required to maintain the required power for max climb angle,...
so in pistons Vx continually increases until it converges with Vy at the absolute performance altitude, whereas a jet produces constant thrust and a constant the AOA is that which produces the greatest climb rate i.e the AOA for L/Dmax
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For max ROC (at VY) we must have maximum excess THP. We will use some of the available THP (the power required) to carry out the work of pushing the aircraft forward against the drag. Any remaining THP available can be used to push the aircraft vertically upwards. The ROC will be this excess THP divided by the weight. We do not necessarily require a high speed. If we had a machine of some kind in which excess power was maximum at zero TAS then that is the speed at which ROC would be maximum(provided we had some means of using that excess power to lift the machine vertically upwards).
This will be true only if the THP available is constant. This is not the case in any aircraft that I am aware of. With any type of prop the thrust decreases with increasing TAS. So the THP curve starts at zero (at zero TAS), increases with increasing TAS up to the TAS at which the rate of decrease of thrust outweighs the rate of increase of TAS. Above this speed the THP decreases. So having minimum drag will not necesarily give you maximum excess THP available.
Maximum excess power (and VY) will occur at VMD only if the thrust is constant at all speeds up to VMD, then decreases markedly beyond VMD.
If thrust remained constant above VMD then we would have the situation used to describe the theoretical behaviour of a Pure Jet in which VY is greater than VMD.
The question being discussed did not ask for VX. It asked for the bast rate of climb speed (VY).
The text that you have quoted apears to be more concerned with how VX and XY vary with changes in altitude.
A piston engine may be a constant BHP machine, but a piston/prop is not a constant THP machine. And in determining best ROC it is the excess THP that matters.
it can only said that at Vy you want the highest forward component of velocity along with a vertical component that is based on the amount of excess THP available for climb,.
and the greatest excess of THP and that would occur at Vimd in which you have the lowest drag there for the lowest thrust required at the highest TAS possible giving the greatest difference between THPr and THPa,...
Maximum excess power (and VY) will occur at VMD only if the thrust is constant at all speeds up to VMD, then decreases markedly beyond VMD.
If thrust remained constant above VMD then we would have the situation used to describe the theoretical behaviour of a Pure Jet in which VY is greater than VMD.
I think the question meant to say Vy not Vx as stated
The question being discussed did not ask for VX. It asked for the bast rate of climb speed (VY).
The text that you have quoted apears to be more concerned with how VX and XY vary with changes in altitude.
Because at that point in a piston [a constant power variable thrust machine--
Since the last one is the easiest I'll start there
that will occur at the highest TAS, because you can't have a speed that gives a lower drag Coefficient in order to increase THP TAS must increase for constant power; of course the power setting will be lower for level flight than climbing flight,..but the maximum excess thrust in the case stated still should occur at Vimd as both thrust and drag decrease together in a climb
that is true because TAS. rho, and AoA i.e induced drag will be changing,..but generally Vy is only one tabulated speed, but constant power implies that the product of T*V remains constant,...although that is an idealized situation
so as you stated
that's not the text, that's one of my boring rants on Pprune
now, I'm not really a morning person so I make no claim that I have not put some horse hooey in the mix,..but I try
The question being discussed did not ask for VX. It asked for the bast rate of climb speed (VY).
For max ROC (at VY) we must have maximum excess THP
but a piston/prop is not a constant THP machine. And in determining best ROC it is the excess THP that matters.
so as you stated
So having minimum drag will not necesarily give you maximum excess THP available.
The text that you have quoted apears to be more concerned with how VX and XY vary with changes in altitude.
now, I'm not really a morning person so I make no claim that I have not put some horse hooey in the mix,..but I try
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that (Max Excess THP) will occur at the highest TAS,
OK let's test that theory. We accelerate to our maximum possible speed where power available = power required and thrust equals drag. We now have zero excess thrust and zero excess THP. So our best climb angle is zero and our best roc is zero.
We will not get best excess THP by flying at max TAS. We will get best excess THP by flying at the speed at which power available minus power required is a maximum. The value of this speed depends on the shape of the power available and power required curves.
Most texts show prop power available as a hump shape starting at zero when TAS equals zero. All power available curves that I have ever seen look a bit like a Nike tick.
I believe that you are an ex RAF fighter pilot. If so you probably have some old copies of the AP3456. If that is the case I suggest you look at them. You will not see a horizontal line for prop power available. If you do not have the 3456 then pretty well any similar text will do. They all show the curves as I have decsribed them.
but constant power implies that the product of T*V remains constant,..
.although that is an idealized situation
If you look at those diagrams in the 3456 (I'm looking at them now) (Yes I stole one as well) you will see that VY prop is more than Vmp, but significantly less than Vmd.
Increasing engine power and adding a constant speed prop will extend the THP curve out the right. This will take Vy closer to Vmd. Now if we go back to the original question we will see that this gives us option c as the correct answer.
TAS must increase for constant power; of course the power setting will be lower for level flight than climbing flight,..but the maximum excess thrust in the case stated still should occur at Vimd as both thrust and drag decrease together in a climb
Keith
I did mean the highest TAS along with the lowest Cd, in the example posted at Vimd the IAS is 125 but if there min drag you will have xs power available I don't mean at a speed where Pr = Pa
basically Vx =[Pa -Pr] max and Vy = [THPa -THPa] max
the hump begins when the curve gets dragged down [the function for the THP is 1/2 rho SC[l,d] V^3] not the drag/thrust curve the THP curve specifically
agreed I did not word that too clearly
the hump occurs/begins when drag increases faster lift
no, not at all ---
I did mean the highest TAS along with the lowest Cd, in the example posted at Vimd the IAS is 125 but if there min drag you will have xs power available I don't mean at a speed where Pr = Pa
basically Vx =[Pa -Pr] max and Vy = [THPa -THPa] max
the hump begins when the curve gets dragged down [the function for the THP is 1/2 rho SC[l,d] V^3] not the drag/thrust curve the THP curve specifically
We will not get best excess THP by flying at max TAS
the hump occurs/begins when drag increases faster lift
I believe that you are an ex RAF fighter pilot
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We clearly are not going to agree on this, so further debate probably won't help.
So we have three views,
Vmp
Between Vmp and Vmd
Vmd
Here's another question taken directly from the JAR CQB (which will not resolve the debate)
Maximum endurance for a piston aeroplane is achieved at?
a. The speed for maximum CL.
b. The speed for minimum drag.
c. The speed that corresponds to the speed for maximum angle of climb.
d. The speed that approximatley corresponds to the speed for maximum rate of climb.
The examiner's answer is option d.
So we have three views,
Vmp
Between Vmp and Vmd
Vmd
Here's another question taken directly from the JAR CQB (which will not resolve the debate)
Maximum endurance for a piston aeroplane is achieved at?
a. The speed for maximum CL.
b. The speed for minimum drag.
c. The speed that corresponds to the speed for maximum angle of climb.
d. The speed that approximatley corresponds to the speed for maximum rate of climb.
The examiner's answer is option d.
I've found some material of help
Embry-Riddle - Aircraft Climb Performance
Pounds thrust to horsepower [Archive] - PPRuNe Forums
Embry-Riddle - Aircraft Climb Performance
Pounds thrust to horsepower [Archive] - PPRuNe Forums
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R.V Davies Classic Might be worth a look now its around £3.
r.v davies | eBay
Its the book that inspired Pete Swatton to pen his
r.v davies | eBay
Its the book that inspired Pete Swatton to pen his
Last edited by cjd_a320; 20th Sep 2011 at 18:52. Reason: forgot the link
