Base turn Distance
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Thanks Batman ... took a while to register and even longer to get to the relevant part of the GEN section, but finally got there. I've read it most thoroughly and can't see any reference to the speed limit being related to terrain clearance. Unless I've missed something, I suspect that you've misconstrued the situation because of the reference in the heading :
"Part III Procedure Construction and Obstacle Clearance Criteria for Instrument Approach Procedures"
This is merely a repetition of the heading used in Doc 8168. It is used there for convenience because Pans Ops is all about obstacle clearance. Whether or not significant obstacles exist in the vicinity of any particular airport or not.
I suspect that the UK CAA's differences need to be read in conjunction with their other publications. That might suggest, as reynoldsno1 said, that the reason is probably more related to tight airspace constraints than to terrain clearance.
Anyway, thanks for the reference. Seems like I could have applied a similar speed restriction here ... sure would be helpful in reducing the size of procedure protection areas!
"Part III Procedure Construction and Obstacle Clearance Criteria for Instrument Approach Procedures"
This is merely a repetition of the heading used in Doc 8168. It is used there for convenience because Pans Ops is all about obstacle clearance. Whether or not significant obstacles exist in the vicinity of any particular airport or not.
I suspect that the UK CAA's differences need to be read in conjunction with their other publications. That might suggest, as reynoldsno1 said, that the reason is probably more related to tight airspace constraints than to terrain clearance.
Anyway, thanks for the reference. Seems like I could have applied a similar speed restriction here ... sure would be helpful in reducing the size of procedure protection areas!
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Base Turns
180 kt 1.8mi
210 kt 2.1mi
Looking at x tk indicator, these work pretty well
210 kt 2.1mi
Looking at x tk indicator, these work pretty well
On a downwind leg at ~330kts at a DME of 8 miles & 5000', 5 miles displaced to the north of the centreline, 15kts of crosswind from the south, I wanted to know how much further I should extend downwind and subsequently what angle of bank I should aim for to ensure I rolled out on the centreline at an appropriate speed/height (i.e., gate) for the concurrent DME?
My limitations are obviously 30 degrees AOB, a radius of turn of 2.5nm and maximum gear extend 270kts whereupon the object of the exercise was to ensure that the average speed for the turn gave the appropriate rate of turn at a constant angle of bank.
In the interests of discussion, any interested parties care to share their thought processes in endeavouring to complete the above visual in minimum track miles?
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Hi chaps
I'm flying often a standard NDB approach. Outbound leg is 1 1/2 min on 037°, then rightturn to inbound track of 241°. this procedure is designed for Cat. B aircraft at 140kts. So this means I shall start my inbound turn after 3.5NM. Does anyone know what the range of the protected area for such a procedure is, or where I have to search for it? Thanks
I'm flying often a standard NDB approach. Outbound leg is 1 1/2 min on 037°, then rightturn to inbound track of 241°. this procedure is designed for Cat. B aircraft at 140kts. So this means I shall start my inbound turn after 3.5NM. Does anyone know what the range of the protected area for such a procedure is, or where I have to search for it? Thanks
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I can't really answer your question without a few more specifics Roastbeef, but it will probably suffice to say that there is protection that includes consideration of a fairly significant tailwind (actually an omni-directional wind), plus an allowance for an error in your outbound timing, plus an allowance for the time it should take you to recognise the point at which you should turn inbound and a further allowance for the time it takes to establish the bank angle and for the plane to start turning.
There's a bit more to it than that, but that's the essence of the primary protection area. Then there's a secondary protection area around the outside of all that, to a distance of 2.5 NM. Within this area, obstacle clearance progerssively reduces to zero, so it's probably not the best idea to test the limits of the protection area too far!
I suspect that the terminology used to describe the approach, "standard", was intended to imply that it's a base turn procedure. This is soemtimes known as a tear-drop turn. I think that he (she?) was trying to differentiate between this type of approach and one that uses a procedure turn at the end of the outbound leg.
There's a bit more to it than that, but that's the essence of the primary protection area. Then there's a secondary protection area around the outside of all that, to a distance of 2.5 NM. Within this area, obstacle clearance progerssively reduces to zero, so it's probably not the best idea to test the limits of the protection area too far!
I suspect that the terminology used to describe the approach, "standard", was intended to imply that it's a base turn procedure. This is soemtimes known as a tear-drop turn. I think that he (she?) was trying to differentiate between this type of approach and one that uses a procedure turn at the end of the outbound leg.
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I'm hijacking the thread back....
FWIW, here is a table of turn radius for a combinations of bank angle (degrees) and TAS (kts):
10 15 20 25 30 35 40 45
150 1.86 1.22 0.90 0.70 0.57 0.47 0.39 0.33
180 2.67 1.76 1.30 1.01 0.82 0.67 0.56 0.47
200 3.30 2.17 1.60 1.25 1.01 0.83 0.69 0.58
250 5.16 3.39 2.50 1.95 1.58 1.30 1.08 0.91
300 7.43 4.89 3.60 2.81 2.27 1.87 1.56 1.31
340 9.54 6.28 4.62 3.61 2.91 2.40 2.01 1.68
From here it is a simple exercise to calculate the instantaneous turn rate whereupon one may verify the reasonableness of the following aide memoire (which I must confess I don't find particularly useful as I think in terms of track miles not time):
Bank Angle for Rate One turn = TAS/10+7
With the above, it is also a simple exercise to calculate track miles in the turn (~=3*turn radius), whereupon with an appropriate figure to hand of the rate of deceleration you are going to effect (clean & level + no speedbrake, clean & level + speedbrake, clean & descending + speedbrake, dirty etc etc), one may calculate the point at which to turn to hit your chosen gate.
FWIW, here is a table of turn radius for a combinations of bank angle (degrees) and TAS (kts):
10 15 20 25 30 35 40 45
150 1.86 1.22 0.90 0.70 0.57 0.47 0.39 0.33
180 2.67 1.76 1.30 1.01 0.82 0.67 0.56 0.47
200 3.30 2.17 1.60 1.25 1.01 0.83 0.69 0.58
250 5.16 3.39 2.50 1.95 1.58 1.30 1.08 0.91
300 7.43 4.89 3.60 2.81 2.27 1.87 1.56 1.31
340 9.54 6.28 4.62 3.61 2.91 2.40 2.01 1.68
From here it is a simple exercise to calculate the instantaneous turn rate whereupon one may verify the reasonableness of the following aide memoire (which I must confess I don't find particularly useful as I think in terms of track miles not time):
Bank Angle for Rate One turn = TAS/10+7
With the above, it is also a simple exercise to calculate track miles in the turn (~=3*turn radius), whereupon with an appropriate figure to hand of the rate of deceleration you are going to effect (clean & level + no speedbrake, clean & level + speedbrake, clean & descending + speedbrake, dirty etc etc), one may calculate the point at which to turn to hit your chosen gate.
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thank you OzExpat,
let's call it a baseturn procedure then
the figure I was looking for is the 2.5NM!
I'm certainly not going to play with the margins, but some people fly this procedure some knots faster then published because of the A/C configuration! And I would like to know where the legal limitations are! Although there is no terrain around within MSA which could hit you (or is it opposite
) I think this is negative training.
Thank's for all the information!
let's call it a baseturn procedure then
the figure I was looking for is the 2.5NM!
I'm certainly not going to play with the margins, but some people fly this procedure some knots faster then published because of the A/C configuration! And I would like to know where the legal limitations are! Although there is no terrain around within MSA which could hit you (or is it opposite
) I think this is negative training. Thank's for all the information!
411A - what complete ar$e of a controller told you to hold at such an absurdly low speed?
Above 34000ft, the restriction is purely 'at or below 0.83TMN'.
It is not for nothing that such idiots have been known for years as the 'Flying Prevention Branch'!
SR71. Radius of turn = (TAS squared)/ (g x tan AoB)
Tables are for simpletons. If you need to be able to work out your turn radius, it's worth doing it properly.
Above 34000ft, the restriction is purely 'at or below 0.83TMN'.
It is not for nothing that such idiots have been known for years as the 'Flying Prevention Branch'!
SR71. Radius of turn = (TAS squared)/ (g x tan AoB)
Tables are for simpletons. If you need to be able to work out your turn radius, it's worth doing it properly.
Last edited by BEagle; 7th Oct 2004 at 17:18.
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Beagle,
The table is worked out using the formula. If you can evaluate the formula during the turn itself, credit where credit is due, but, me, being a neanderthal, I need a simple aide memoire hence the table...
The rate of turn is:
d/dt(\gamma)=V/r=g*sqrt(n^2-1)/V
where n = load factor.
n is obviously a function of the AOB.
I realised shooting the above approach I had an aide memoire for decel rate, track miles to go below FL100 but not one for turn radii at a particular speed/AOB combination...
PS: John T - Any update on getting [tex] tag support?
The table is worked out using the formula. If you can evaluate the formula during the turn itself, credit where credit is due, but, me, being a neanderthal, I need a simple aide memoire hence the table...
The rate of turn is:
d/dt(\gamma)=V/r=g*sqrt(n^2-1)/V
where n = load factor.
n is obviously a function of the AOB.
I realised shooting the above approach I had an aide memoire for decel rate, track miles to go below FL100 but not one for turn radii at a particular speed/AOB combination...
PS: John T - Any update on getting [tex] tag support?
Why on earth try to solve a differential equation whilst flying an aeroplane?
What is it you need to know? The time it'll take to turn through a given number of degrees at a given TAS and AOB?
In which case if you want to turn through n deg, the time taken will be n x (2pi x TAS x 60)/( 360 x g x tan AoB) seconds.
At 240 TAS and 25 deg AoB, that simplifies to roughly n/2 seconds...or 0.4712 n if you prefer.
About 2/3 Rate 1
What is it you need to know? The time it'll take to turn through a given number of degrees at a given TAS and AOB?
In which case if you want to turn through n deg, the time taken will be n x (2pi x TAS x 60)/( 360 x g x tan AoB) seconds.
At 240 TAS and 25 deg AoB, that simplifies to roughly n/2 seconds...or 0.4712 n if you prefer.
About 2/3 Rate 1
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I asked the following a number of posts ago:
Thats what I wanted to know...
I'm quite aware of all the formulas but the point was, in the absence of an applicable aide memoire to calculate the turn radius, bearing in mind I'd rather be looking out the window on a visual than punching numbers into my calculator (as I hazard a guess you still did to evaluate the formula in your previous post? If not, you obviously have some aide memoire for calculating \tan(\phi)...), I realised I wasn't sure if my lateral displacement from the centreline was enough to ensure I didn't travel through it whilst making the turn...
Of course, in reality the differential equations do need to be solved in situ because I chose to decelerate and change the configuration in the turn. That was the whole point of the exercise in terms of reducing the track miles of the visual to the minimum by using all the appropriate aides available.
The exercise then became one of approximation...
If I started my turn at 330kts and was endeavouring to complete it at 180kts, is it geometrically possible to do so within 5 miles? (I chose to approximate such a turn by using an average of 250kts, whereupon, your/my calculations show that one only needs a lateral displacement from the centreline of 4 miles to turn through 180 degrees....even less with a crosswind in the favourable direction. Whereupon, one may relax the bank angle towards the end of the turn to roll out on the correct inbound track.)
If so, is it also possible to loose 1000' and be in the slot?
If not, what are my options?
Shall I extend further downwind and decelerate before making the turn?
So, in short, I wanted to know quite a lot of things...
In the interests of discussion, any interested parties care to share their thought processes in endeavouring to complete the above visual in minimum track miles?
I'm quite aware of all the formulas but the point was, in the absence of an applicable aide memoire to calculate the turn radius, bearing in mind I'd rather be looking out the window on a visual than punching numbers into my calculator (as I hazard a guess you still did to evaluate the formula in your previous post? If not, you obviously have some aide memoire for calculating \tan(\phi)...), I realised I wasn't sure if my lateral displacement from the centreline was enough to ensure I didn't travel through it whilst making the turn...
Of course, in reality the differential equations do need to be solved in situ because I chose to decelerate and change the configuration in the turn. That was the whole point of the exercise in terms of reducing the track miles of the visual to the minimum by using all the appropriate aides available.
The exercise then became one of approximation...
If I started my turn at 330kts and was endeavouring to complete it at 180kts, is it geometrically possible to do so within 5 miles? (I chose to approximate such a turn by using an average of 250kts, whereupon, your/my calculations show that one only needs a lateral displacement from the centreline of 4 miles to turn through 180 degrees....even less with a crosswind in the favourable direction. Whereupon, one may relax the bank angle towards the end of the turn to roll out on the correct inbound track.)
If so, is it also possible to loose 1000' and be in the slot?
If not, what are my options?
Shall I extend further downwind and decelerate before making the turn?
So, in short, I wanted to know quite a lot of things...
PPRuNeaholic
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I might be missing something here SR71, but here's my simplistic idea. If you start the turn 4 NM prior to the required inbound track at a speed of about 4 miles per minute, the turn will take about a minute. Thus, if you want to lose 1,000 feet during the manoeuvre, you need a ROD of about 1,000 feet per minute.
In practice, that will work out close enough IF you can achieve an average of 1,000 FPM descent through the turn without popping pax ear drums. I assume you're flying a pressurised aeroplane so there shouldn't be a problem as long as cabin diff doesn't get excessive.
In practice, that will work out close enough IF you can achieve an average of 1,000 FPM descent through the turn without popping pax ear drums. I assume you're flying a pressurised aeroplane so there shouldn't be a problem as long as cabin diff doesn't get excessive.
Is this an attempt by the world's least favourite lo-co to work out some cowboy procedure to shave another 100kg off their sector fuel burn?
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Personally i just fly the plate at the speed i'm supposed to and read any special brief for the proceedure. As for the need to judge height over trackmiles ,well just how do you quantify an art form as a formula.
dont over complicate your life, look at the prog page (if youve got one ) track miles. If not artform applies.
Once on the gnd i tend to read the paper after i've calculated the radius of my walk around thus:
rain +1f/o= waving@drowned rat x speed of walk
So much theory and so few pilots.
dont over complicate your life, look at the prog page (if youve got one ) track miles. If not artform applies.
Once on the gnd i tend to read the paper after i've calculated the radius of my walk around thus:
rain +1f/o= waving@drowned rat x speed of walk
So much theory and so few pilots.
