performance question 4

[gphillips134]

what rate of climb will be achieved with a climb gradient of 3.3% and a groundspeed of 100kts?

explaination thank you

[Keith.Williams.]

If you go to page 28 of the CAP 698 you will find the equation:

Still air gradient % = ROC/TAS x 6000/6080

You want the ROC, so rearrange the equation by multiplying both sides of the equation by (TAS x 6080), then divide both sides by 6000.

This will give the equation:

ROC = (Still air gradient x TAS x 6080) / 6000

Putting in the numbers given in the question gives:

ROC = (3.3 x 100 x 6080) / 6000

Which is ROC = 334.4 ft/min

Alternatively you could simply remember that 1 knot = 101.3 ft/min. For small angles of climb the ROC is approximately the % gradient x the TAS. In this case we have 3.3% x 101.3 ft/min, which is ROC = 334.4 ft/min. This method actually uses the same equation but does not express it explicitly.

But for most students it is easier to forget about memorising equations and simply refer to the CAP. The advice given by Pugzi when answering your previous question was very true. You can get lots of marks in the Performance exam simply by knowing your way around the CAP 698.

[gphillips134]

thanks very much for the explaination to the question and to the people that replied before

[keeno]

hey there,

there is a much simpler way of doing it...


% = (FPM/TAS) x 0.987

If you re arange the equation so that you get

FPM = (TASx%) / 0.987

you will get the same answer of 334.3 as your rate of climb...

I find this way much easier..works all the time

Keeno

[Keith.Williams.]

KEENO,

The reason your equation works is because it is the same as the one in the CAP698. 0.987 is approximatley 6000/6080.

And 101.3 (ft/min per Knot) is approximately 1/0.987, which is why all three methods work. They are all just different versions of the same equation.

But you are correct, yours (appears to be) easier.