What's a 12% glideslope? (asks a mentally retarded ATPL student)

[Gin Slinger]

What is meant by a 12% glideslope?

Is it 12% of a 90 deg from horizontal (a vertical 'glideslope'), i.e. 10.8 deg?

Or is it 12% of a 1-to-1 glideslope (45% angle from horizontal/vertical), i.e. 5.4 deg?

Or is it a twelve-along-one-down affair, ie ~7.2 deg from hortizontal?

[sorry for the stupid questions, I've confused the hell out of myself with this one!]

[Gin Slinger]

...and another one....

I'm cruising at 450kt TAS with a +50 wind component.

What is the ground distance equivalent to 760 NAM?

I reckon 854nm - ([500/450] x 760) , but the answer given is 780nm

Please help! :confused:

[Keith.Williams.]

Gin,

a 12% glideslope means that for every 100 ft you move down the slope, you move 12 ft vertically downwards

To find the angle just remember that 12% = 0.12 an dthis is the sine of the glideslope angle.

This is because the hypoteneuse is the path you have taken and the opposite is the vertical distance you have moved downwards.
So the sine is 12/100 or 0.12.

So a 12% glideslope means going down a slope of 6.89 degrees below the horizontal.

Climb gradients are the same but (obviously) upwards!!

All of this assumes you are in still air.

[Babi Melayu]

Please don't call yourself a retard.


You might be familiar with the more common 3-degree glideslope or gradient of 5%. The 5% means 5 / 100, 5 feet lost vertically (y-axis) for every 100 feet travelled horizontally (x-axis). When you intend to find the angle subtended by this 5 (y-axis) and 100 (x-axis), you just take the inverse tangent of them both - opposite over adjacent i.e. inverse tangent of 5 / 100. The result is an angle of 2.83 degrees ( which is close to the 3-degree we mentioned).


The TWELVE percent (12%) slope means you lose 12 feet vertically for every 100 feet travelled horizontally and this is simply a much steeper slope i.e. a whopping 6.84 degrees glideslope! ( Inverse tangent of 12 over 100 gives us 6.84)


Who uses the 12% slope, by the way?

[Gin Slinger]

Keith & Babi - thanks!

Babi: it's just a question that's come up in one of my JAA ATPL Gen Nav practice paper.

Can anyone help with the NAM question?

[Keith.Williams.]

Gin,

wrt your second question, I think that you are correct (ish)

500/450 x 760 is actually 844.4 not the figure you quoted.

It is possible that your quoted (correct) answer is wrong.
It happens all the time with feedback from the ATPL exams.

wrt the glideslope problem, tan for small angles almost = sine

I'm note sure that tan is any easier, but just use whichever one you prefer.

But for any obstacle clearance questions it is worth noting that for small climb gradients, the height gained is approximately equal to the ground distance to the obstacle multiplied by the % gradient as a decimal fraction.

So for example, over a 1000m range to an obstacle in a 12% climb ragient you gain about 1000 x 0.12 = 120 m of height.
(or about 393.6 ft)

[OneDotLow]

Are you sure you are not getting confused with a 12 degree glideslope... ie the false glideslope that occurs at around 12 degrees?
It is an unwanted biproduct of the standard 3 deg glideslope; sorry i dont remember what causes it.
Cheers
ODL

[Gin Slinger]

Thanks again Keith - yes, 844.4 - the 850 was from memory.

Regarding tan/sine usage, I must admit, just using the GCSE Maths trig numonic (SOH CAH TOA) is easiest for me.