Q1
In an example in lesson 13 glidepath is given as 12% and the degrees are required. Calculation goes 12x60+100=7.2 degrees. Why does the calculation multiply percent with 60+100 to get degrees when therelationship is actually a function of tangent?
Q2
I truly can't figure out how to get the ROD with CRP5 following the given instructions. The photo provided shows outer, fixed, temperature conversion scale encircled in green. What does it have to do with ROD?
The way to get 700ft ROD with CRP5 would be using multiplication ratio  set a GS over 10, and read 700 over a multiplier of 5 (70 over 50). What am I doing wrong?
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Lesson 13  Glidepath calculations percentage to degrees
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Lesson 13  Glidepath calculations percentage to degrees
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All clear now, thanks JJ Jones! I had a short brake and moved forward with the syllabus, just now stepped by here. Cheers!

For Q1
You are correct  these are a function of the tangent relationship.
However we are required to use rules of thumb for this part of the syllabus. Using the CRP5 or the "rule of thumb" formula you would get a 12 percent gradient equating to a 7.2 degree glideslope. (12 times 60 divided by 100 is 7.2).
For a closer approximation, take the gradient in % and multiply by .57 . For example, 12 X .57 = 6.84 degrees. .
To be exact, inv tan(gradient) = climb angle.
Inv tan(.12) = 6.84277 degrees
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I'll take Q2 first
The diagram on page 4 of 15 is the wrong one. I will get it changed.
In the interim here is an example of the use of the CRP5 to work out the ROD. 3 degree glide against the black triangle, 140 Knots groundspeed on the inner scale and read 700 feet per minute ROD on the outer scale.
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