Question on Turn
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Question on Turn
What's the definition of Instantaneous Turn Rate and Sustained Turn Rate?
I did search google for answer but nothing available.
If there is no fixed definition, is it possible that you answer this by your experience or even your instinct?
I did search google for answer but nothing available.
If there is no fixed definition, is it possible that you answer this by your experience or even your instinct?
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One definition might be - and I'm not going to claim this is definitive - that (Maximum) Instantaneous Turn Rate is the highest rate of turn 9or, if you prefer, load factor) that can be achieved for the speed, weight, configuration, etc. under discussion but without regard for the energy state of the aircraft, whereas (maximum) Sustained Turn Rate requires an additional constraint that the aircraft be able to maintain a stable energy state i.e. the engines be capable of maintaining the speed.
I've seen that differentiation applied to military aircraft turn capability, where the instantaneous rate is likely limited only by structural considerations, certainly at higher speeds, but the sustained rate can be very limited by thrust available.
I've seen that differentiation applied to military aircraft turn capability, where the instantaneous rate is likely limited only by structural considerations, certainly at higher speeds, but the sustained rate can be very limited by thrust available.
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I'm sorry, not clear yet for me.
If STR limited by speed, why not for ITR?
According to my knowledge, structure limitation for maneuver G limit is much higher than ITR. Therefor, something else being determined ITR.
If STR limited by speed, why not for ITR?
According to my knowledge, structure limitation for maneuver G limit is much higher than ITR. Therefor, something else being determined ITR.
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Well, without knowing which aircraft its hard to guess....
I've seen (again in the military context) a distinction being made between the initial turn rate (number of degrees of heading change in first second OR time to achieve first XX degrees) against the rate achievable once established in a turn. In that case the initial turn rate is reduced below the sustained rate by the need to roll into the turn first.
But whether the "instantaneous" you are seeing is what I'd think of as "initial"...?
Perhaps quote the numbers you have and the conditions and aircraft they apply to and someone might be able to reverse engineer the specific definition?
I've seen (again in the military context) a distinction being made between the initial turn rate (number of degrees of heading change in first second OR time to achieve first XX degrees) against the rate achievable once established in a turn. In that case the initial turn rate is reduced below the sustained rate by the need to roll into the turn first.
But whether the "instantaneous" you are seeing is what I'd think of as "initial"...?
Perhaps quote the numbers you have and the conditions and aircraft they apply to and someone might be able to reverse engineer the specific definition?
M.FS,or Rousseau,I can send you some perf.graphs of the L-39 Turning performance,if you let me have your e.mail by PM, as i`m incompetent to download here. ITR is governed by speed/power and structural limit.STR is governed by speed/power and lift(stalling)limit.
M.FS, emails for you..
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OK here's two charts courtesy of sycamore
Looking at the first one, it shows turn perf for aircraft at three different weights.
Look at the innermost of the bold black lines. It starts at just under 200km/h at 0 turn rate - thats the 1'g' stall speed. As you go 'up' the page (and so cross lines of 'g' or turn rate) it curves to the right, showing an increase in speed - that's the stall speed going up as the 'g' increases. That's why the first curve, that bends up and right, is marked 'lift limit'.
Follow that curve all the way up (past the branch at 13.5 deg/sec, which we'll come back to) and you get to a fairly right angled corner, at 7'g', 24 deg/sec, 400m radius. Thats the point where our lift limited 'g' runs into a structural 7'g' limit. So past this point we can't continue to pull to the lift limit - now we follow a line of constant 'g', until we get to 15 deg/sec and 910kph - now we've hit the max speed, and can't go any faster.
that line just followed is the absolute maximum turn performance at 4,400kg - so it represents the maximum instantaneous turn rate.
The reason it is "instantaneous" is that it cannot be achieved for more than a split second.
If we now start again, but this time stop at the branch ignored earlier, at 13.5 deg/sec, and follow the line marked 'sustained turn rate' we are now seeing that the turn rate we can achieve is much lower - at 500kph, for example, the max instantaneous rate is about 22 deg/sec, but the max sustained is about 12 deg/sec. that's because the aircraft doen't have the power to overcome the huge induced drag at 5.5'g', but does have the oomph for about 3'g'. At the max speed end everything collapses to a speed limit of about 700kph - the aircraft hasnt got the power to reach its max speed limit in level flight.
The second chart tells much the same story, except its a very high drag index, so lots of stores. So the aircraft, this time, can't get to 700kph - this time it maxes out at about 650kph.
Does that help at all?
Looking at the first one, it shows turn perf for aircraft at three different weights.
Look at the innermost of the bold black lines. It starts at just under 200km/h at 0 turn rate - thats the 1'g' stall speed. As you go 'up' the page (and so cross lines of 'g' or turn rate) it curves to the right, showing an increase in speed - that's the stall speed going up as the 'g' increases. That's why the first curve, that bends up and right, is marked 'lift limit'.
Follow that curve all the way up (past the branch at 13.5 deg/sec, which we'll come back to) and you get to a fairly right angled corner, at 7'g', 24 deg/sec, 400m radius. Thats the point where our lift limited 'g' runs into a structural 7'g' limit. So past this point we can't continue to pull to the lift limit - now we follow a line of constant 'g', until we get to 15 deg/sec and 910kph - now we've hit the max speed, and can't go any faster.
that line just followed is the absolute maximum turn performance at 4,400kg - so it represents the maximum instantaneous turn rate.
The reason it is "instantaneous" is that it cannot be achieved for more than a split second.
If we now start again, but this time stop at the branch ignored earlier, at 13.5 deg/sec, and follow the line marked 'sustained turn rate' we are now seeing that the turn rate we can achieve is much lower - at 500kph, for example, the max instantaneous rate is about 22 deg/sec, but the max sustained is about 12 deg/sec. that's because the aircraft doen't have the power to overcome the huge induced drag at 5.5'g', but does have the oomph for about 3'g'. At the max speed end everything collapses to a speed limit of about 700kph - the aircraft hasnt got the power to reach its max speed limit in level flight.
The second chart tells much the same story, except its a very high drag index, so lots of stores. So the aircraft, this time, can't get to 700kph - this time it maxes out at about 650kph.
Does that help at all?
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Sorry for long time.
I don't get it.
Clearly the STR reach the G limit also, see the graph below
And from your words, my understanding is ITR determined by Speed whereas STR determined by lift?
I don't get it.
Clearly the STR reach the G limit also, see the graph below
And from your words, my understanding is ITR determined by Speed whereas STR determined by lift?
Do a Hover - it avoids G
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What determines the ITR and the STR is very simple. But you must go back to basics.
Both depend on the lift equation. The ITR ignores the drag side of life while the STR takes drag into account.
The lift equation has all that you need for a complete understanding of both:
L = CL1/2 rSV2
(sorry PPRuNe does not like the symbols. If you don't know the lift eqation then start by finding that out but here r stands for the air density and the V is a squared term)
The max ITR is what you get if you suddenly yank the stick back until you reach Cl max and stall the wing. In doing this the G meter will show some maximum value. Exactly what rate of turn that G value will provide is down to the laws of physics not aerodynamics so let us not get diverted into that other than to say that for a given G value the rate of turn will be highest at the lowest forward speed at which you can achieve that G.
As to how the G you can yank varies with all the other factors just go back to the terms in the lift equation. More density more G. More wing area more G. More V more G and naturally more weight less G. But watch the V term. As mach number is increased it is certain that the Cl max you get in your yank to the stall will decrease due to compressibility effects.
IF (and it is a big if) you now really feel happy with this appreciation of ITR then (and only then) you are ready to consider the STR.
The STR introduces drag effects. To sustain a turn rate you must have enough thrust to overcome the drag increase that will follow your yank to the ITR. Since you could well do your yank when you are ALREADY at full throttle you will quickly slow down and so will then be able to yank to less G. There will come a time where your thrust equals the increased drag from yanking and bingo you can sustain that reduced G.
Perhaps you can now see why in the test flying word the ITR is called the lift boundary and the STR is called the thrust boundary.
So in conclusion give a particular aircraft more thrust and you will increase the STR or thrust boundary. The effect of more thrust on the ITR will be negligible.
PS The aircraft's (structural) G limit will naturally limit any manoeuvres should it intersect the curves of ITR (very likely at high speed) or STR (possible with a VERY hiigh thrust to weight aircraft).
Both depend on the lift equation. The ITR ignores the drag side of life while the STR takes drag into account.
The lift equation has all that you need for a complete understanding of both:
L = CL1/2 rSV2
(sorry PPRuNe does not like the symbols. If you don't know the lift eqation then start by finding that out but here r stands for the air density and the V is a squared term)
The max ITR is what you get if you suddenly yank the stick back until you reach Cl max and stall the wing. In doing this the G meter will show some maximum value. Exactly what rate of turn that G value will provide is down to the laws of physics not aerodynamics so let us not get diverted into that other than to say that for a given G value the rate of turn will be highest at the lowest forward speed at which you can achieve that G.
As to how the G you can yank varies with all the other factors just go back to the terms in the lift equation. More density more G. More wing area more G. More V more G and naturally more weight less G. But watch the V term. As mach number is increased it is certain that the Cl max you get in your yank to the stall will decrease due to compressibility effects.
IF (and it is a big if) you now really feel happy with this appreciation of ITR then (and only then) you are ready to consider the STR.
The STR introduces drag effects. To sustain a turn rate you must have enough thrust to overcome the drag increase that will follow your yank to the ITR. Since you could well do your yank when you are ALREADY at full throttle you will quickly slow down and so will then be able to yank to less G. There will come a time where your thrust equals the increased drag from yanking and bingo you can sustain that reduced G.
Perhaps you can now see why in the test flying word the ITR is called the lift boundary and the STR is called the thrust boundary.
So in conclusion give a particular aircraft more thrust and you will increase the STR or thrust boundary. The effect of more thrust on the ITR will be negligible.
PS The aircraft's (structural) G limit will naturally limit any manoeuvres should it intersect the curves of ITR (very likely at high speed) or STR (possible with a VERY hiigh thrust to weight aircraft).
Last edited by John Farley; 15th Oct 2011 at 13:08.
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Not forgetting, of course, the poor old 'human' part of the aircraft who has his or her Instantaneous and Sustained 'g' limit.
Just for old times sake, the dear old EE/BAC Lightning could, at lowish weights, sustain 6g from sea level up to around 10k (in reheat). Not bad for an oldie (mind you, it was out of fuel then
)
Just for old times sake, the dear old EE/BAC Lightning could, at lowish weights, sustain 6g from sea level up to around 10k (in reheat). Not bad for an oldie (mind you, it was out of fuel then
)
