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- Cretella
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Hi,

I can't seem to work out where I'm going wrong with regaining FPT in point marked X in 1 in 60 questions.

I've been looking in RPL/PPL NAV section which shows you calculate CA with the same formula as TE (dx60)/D and TTI is the sum of TE and CA. In the questions I've been doing using the formula hasn't been working

Instead of working out working out TE immediately I worked out the difference between 1st Point and 2nd Point because the first point wasn't on the same line like the questions in the book then applied the TE formula and worked out the second one normally.

Please help

Thank you

I can't seem to work out where I'm going wrong with regaining FPT in point marked X in 1 in 60 questions.

I've been looking in RPL/PPL NAV section which shows you calculate CA with the same formula as TE (dx60)/D and TTI is the sum of TE and CA. In the questions I've been doing using the formula hasn't been working

Instead of working out working out TE immediately I worked out the difference between 1st Point and 2nd Point because the first point wasn't on the same line like the questions in the book then applied the TE formula and worked out the second one normally.

Please help

Thank you

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- John.Heddles
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- ATPL/consulting aero engineer

- Posts: 697
- Thank you received: 76

First, may I suggest steer clear of going straight to a formula to solve these sorts of questions. Far too easy to make a mistake and deny yourself an easy way of picking it up. An absolutely sacrosanct principle in flying is to leave yourself ways of checking solutions and always to make those checks. It is very embarrassing should you get, say, your fuel calcs wrong and not pick the error up until the noise stops. Have a read of the Gimli glider prang to drive this point home ....

So, what to do ?

Always draw the triangles which constitute the problem and its solution - only takes a few seconds and is money in the bank. At the end of the day, 1 in 60 problems fall out super simple once you can see the story right there in front of you. You deny yourself an easy solution process if you try to avoid the few seconds to draw the picture. Am I getting boring by belabouring that point ?

I'm on the wrong laptop so I can't easily draw a picture to post, so let's talk it up.

For the initial triangle, you have flown 15nm and slid off track direction 3nm to the left, so draw that and then the perpendicular to the track direction. Simple back of a fag packet sum to figure that the initial TE is 12 deg to the left. (This initial sum is considering that X is 6nm right of where it is). So, if we turn right 12 degrees, we should (more or less) parallel the required track drawn in the question.

For the second triangle, you have 45nm to go and we need to slide a further 3nm to the left. Now, you can do it the hard way or note that 45/15 = 3/1 so the CA has to be 1/3 of the TE which is 12/3 = 4 (spotting where you can use simple mental arithmetic - and then using those simple techniques - is gold bars in the safe). So you need 4 deg left correction to get (more or less) back on track at X. That is, overall, in your mind you turn 12 deg right to parallel but then 4 deg left to intercept. Overall, thatâ€™s a heading change of 8 deg to the right.

Notice that I have figured two calculations - a heading change to parallel track intended direction, and a further correction to intercept track, or whatever is required. Please don't try to figure it all out in one go - that will make for heartache.

You now have the two triangles in front of you so you can have a looksee to make sure that everything looks fine... and, again, be wary of "canned" solutions. If you get everything right, fine, but it's really very easy to make mistakes and get no marks for the question.

Now, before folks start complaining that I'm being a bit rough and ready and not sufficiently accurate with mental arithmetic, do keep in mind that the 1 in 60 "rule" is not the 1 in 60 rule at all. It is a really useful rough and ready approximation based on simple geometry and trigonometry which shows that, for around 57.3 distance units, a track error of 1 distance unit is around a degree track error. We use 60, (rather than 57.2957795......) because it's a nice round sort of number which (to our sexagesimal trained minds) makes for reasonably easy mental calculations, if you have practiced that a little.

Also, keep in mind that the trigonometry shows that the approximation only works reasonably for small angles up to, say 15-20 degrees. Beyond that, the error rapidly gets a bit out of control.

So, what to do ?

Always draw the triangles which constitute the problem and its solution - only takes a few seconds and is money in the bank. At the end of the day, 1 in 60 problems fall out super simple once you can see the story right there in front of you. You deny yourself an easy solution process if you try to avoid the few seconds to draw the picture. Am I getting boring by belabouring that point ?

I'm on the wrong laptop so I can't easily draw a picture to post, so let's talk it up.

For the initial triangle, you have flown 15nm and slid off track direction 3nm to the left, so draw that and then the perpendicular to the track direction. Simple back of a fag packet sum to figure that the initial TE is 12 deg to the left. (This initial sum is considering that X is 6nm right of where it is). So, if we turn right 12 degrees, we should (more or less) parallel the required track drawn in the question.

For the second triangle, you have 45nm to go and we need to slide a further 3nm to the left. Now, you can do it the hard way or note that 45/15 = 3/1 so the CA has to be 1/3 of the TE which is 12/3 = 4 (spotting where you can use simple mental arithmetic - and then using those simple techniques - is gold bars in the safe). So you need 4 deg left correction to get (more or less) back on track at X. That is, overall, in your mind you turn 12 deg right to parallel but then 4 deg left to intercept. Overall, thatâ€™s a heading change of 8 deg to the right.

Notice that I have figured two calculations - a heading change to parallel track intended direction, and a further correction to intercept track, or whatever is required. Please don't try to figure it all out in one go - that will make for heartache.

You now have the two triangles in front of you so you can have a looksee to make sure that everything looks fine... and, again, be wary of "canned" solutions. If you get everything right, fine, but it's really very easy to make mistakes and get no marks for the question.

Now, before folks start complaining that I'm being a bit rough and ready and not sufficiently accurate with mental arithmetic, do keep in mind that the 1 in 60 "rule" is not the 1 in 60 rule at all. It is a really useful rough and ready approximation based on simple geometry and trigonometry which shows that, for around 57.3 distance units, a track error of 1 distance unit is around a degree track error. We use 60, (rather than 57.2957795......) because it's a nice round sort of number which (to our sexagesimal trained minds) makes for reasonably easy mental calculations, if you have practiced that a little.

Also, keep in mind that the trigonometry shows that the approximation only works reasonably for small angles up to, say 15-20 degrees. Beyond that, the error rapidly gets a bit out of control.

Engineering specialist in aircraft performance and weight control.

Last Edit: 1 month 3 weeks ago by John.Heddles.

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- John.Heddles
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- ATPL/consulting aero engineer

- Posts: 697
- Thank you received: 76

You made reference to questions, plural, so there may be a systemic (ie understanding) problem there. Invariably, such problems aren't the province of one student and there will be others who are struggling with similar woes.

Perhaps you could post, say, several of the particular questions with which you are having problems and we can post, in return, detailed worked answers (with sketches) which probably will assist a number of folk ?

Engineering specialist in aircraft performance and weight control.

Last Edit: 1 month 2 weeks ago by John.Heddles.

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