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- Mister W

Last edit: 9 years 8 months ago by Mister W.

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- Nikita

One more question to the experts:

When doing 1-in-60 exercises I am having a problem to calculate the New HDG after Intercept (M).

For example:

Given

**FPT= 273 M, Heading held= 274 M, Distance along Track= 80 nm, Distance off Track = 13 nm left, Distance to re-join Track= 50 nm.**

I have found Track Error= 10 , TMG = 263 , Drift = 11 left, Closing angle = 16, TTI = 289, and HDG to Intercept = 300 M.

I just missing the step of how to calculate the new HDG after Intercept.

Cheers,

Nikita

When doing 1-in-60 exercises I am having a problem to calculate the New HDG after Intercept (M).

For example:

Given

I have found Track Error= 10 , TMG = 263 , Drift = 11 left, Closing angle = 16, TTI = 289, and HDG to Intercept = 300 M.

I just missing the step of how to calculate the new HDG after Intercept.

Cheers,

Nikita

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- Mister W

Hi Nikita,

Now that you have calculated the drift, 11° in this example, you can now apply it to your FPT. This will give you a new HDG which should keep you on track much better than what you originally planned. So we end up with planned track of 273° + 11° to the right = New HDG 284°.

Always found DR nav to be the most fun to do and it has saved my neck more than once when things like GPS and navaids failed.

Hope this helps.

Cheers,

Mister W.

Now that you have calculated the drift, 11° in this example, you can now apply it to your FPT. This will give you a new HDG which should keep you on track much better than what you originally planned. So we end up with planned track of 273° + 11° to the right = New HDG 284°.

Always found DR nav to be the most fun to do and it has saved my neck more than once when things like GPS and navaids failed.

Hope this helps.

Cheers,

Mister W.

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- Nikita

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- abargri

Hello all,

My first post ever - and hope that there will be many more.

Just on the Pressure Height calculation, I got 13,740. That is 1013-1015 = -2x30 = 60 and 13800-60 0= 13740

How did we get 13510? What am I missing?

cheers

My first post ever - and hope that there will be many more.

Just on the Pressure Height calculation, I got 13,740. That is 1013-1015 = -2x30 = 60 and 13800-60 0= 13740

How did we get 13510? What am I missing?

cheers

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- Mister W

I think QNH 1015 was for the question below the one we're discussing. You had me thinking though!

Cheers,

Mister W.

Cheers,

Mister W.

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- abargri

I'm not sure M W... I saw 1015 in the very first post, so now I'm really confused :-S

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- Richard
- Topic Author

Hi Abargri,

The actual question in the workbook is Alt: 13800 ft, QNH 1022 and OAT 22 degrees C. This gives the Pressure altitude of 13530 ft. There was a misprint in the first post which mentioned the QNH being 1015. I have edited Nikita's original post to make the QNH 1022.

If the QNH was indeed 1015, your calculation would be correct

Cheers,

Rich

The actual question in the workbook is Alt: 13800 ft, QNH 1022 and OAT 22 degrees C. This gives the Pressure altitude of 13530 ft. There was a misprint in the first post which mentioned the QNH being 1015. I have edited Nikita's original post to make the QNH 1022.

If the QNH was indeed 1015, your calculation would be correct

Cheers,

Rich

Last edit: 9 years 6 months ago by Richard.

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- JamesScot

I thought that we weren't supposed to round until the final answer?

i.e., ISA = 15- (2*13530/1000) = -12.06

ISA Dev = 22- -12.06 = 34.06 degrees

Add elevation 34.06 *120 = 4087.2

Density Altitude = 17617ft

This particular one isn't massively out. But the first one is

My answer

FE 4440ft / QNH 999hPa / OAT15 degrees => PH = 4840ft / ISA TEMP 5.32 / ISA DEV 9.68 / Add alt 1161.6 / =>>> DH=5602ft

But the answer on that sheet is 6060ft. A massive 460 ft difference in density height.

This is entirely due to the rounding of the pressure height before calculations start.

Is that an acceptable error difference? How does one know when to round the PH? Does it go up in 500ft or 250ft increments or 100ft increments?

What about the temperature deviation? Bear in mind, that a 0.5 degree dev from ISA results in a 60ft difference. If you round to the nearest 100ft for pressure height and then to the nearest whole degree for ISA temp deviation (that's 100ft + 0.5*120ft = 110ft) is a 110ft deviation from actual DH. If you round pressure height to the nearest 500ft and the nearest whole degree for ISA temp deviation (500ft + 0.5*120)=560ft deviation from actual DH.

I didn't think that level of accuracy/inaccuracy was acceptable. I thought we weren't supposed to round until the very end.

Help!

i.e., ISA = 15- (2*13530/1000) = -12.06

ISA Dev = 22- -12.06 = 34.06 degrees

Add elevation 34.06 *120 = 4087.2

Density Altitude = 17617ft

This particular one isn't massively out. But the first one is

My answer

FE 4440ft / QNH 999hPa / OAT15 degrees => PH = 4840ft / ISA TEMP 5.32 / ISA DEV 9.68 / Add alt 1161.6 / =>>> DH=5602ft

But the answer on that sheet is 6060ft. A massive 460 ft difference in density height.

This is entirely due to the rounding of the pressure height before calculations start.

Is that an acceptable error difference? How does one know when to round the PH? Does it go up in 500ft or 250ft increments or 100ft increments?

What about the temperature deviation? Bear in mind, that a 0.5 degree dev from ISA results in a 60ft difference. If you round to the nearest 100ft for pressure height and then to the nearest whole degree for ISA temp deviation (that's 100ft + 0.5*120ft = 110ft) is a 110ft deviation from actual DH. If you round pressure height to the nearest 500ft and the nearest whole degree for ISA temp deviation (500ft + 0.5*120)=560ft deviation from actual DH.

I didn't think that level of accuracy/inaccuracy was acceptable. I thought we weren't supposed to round until the very end.

Help!

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- John.Heddles
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- ATPL/consulting aero engineer

- Posts: 876
- Thank you received: 108

As a general rule approach to things, that's a good idea, noting that, for exam questions, the examiner may use words which vary any generally expected technique.

Things get a tad messy with atmospheric hydrostatics. The equations are not linear (generally exponential - and these, themselves, are only theoretical approximations to the real world) and the various approximations we use in aviation are just that, and pretty rough and ready at best. Although we use approximations such as 30 ft/hPa, 120 ft/C°, for example, such numbers are very rubbery, especially for pressure and density variations.

The end result is that the answers are fine for the exam as the aim is to check that the candidate has an idea of the story but, in reality, they don't represent anything particularly accurate in the real atmosphere - they are, at best, useful indications of what's going on. You want the correct answer, you need to use the correct equation - which is rather somewhat beyond what would be reasonable for pilot training.

The takeaway is that one ought not to get too fussed about super accuracy in altimetry exam question matters ....

Engineering specialist in aircraft performance and weight control.

Last edit: 3 years 3 months ago by John.Heddles.

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