I have been calculating this question but I just don't seem to get the correct answer.

For this question use Performance Manual CAP 698 SEP1 Figure 2.3.

Using the climb performance chart, for the single-engine aeroplane, determine the ground distance to reach a height of 2000 ft in the following conditions:


OAT at take-off: 25°C

Airport pressure altitude: 1000 ft

Aeroplane mass: 3600 lb

Speed: 100 KIAS

Wind component: 15 kt Headwind

My Workings

*TAS = 104kt (from using CRP-5)

GS = 104kts - 15kts = 89kts* Climb Gradient = 10.7% (from chart)

Still Air Distance = (2000 / 10.7) X 100 = 18,691ft

Ground Distance = 18691 X (89/104) = 15,995ft

But the CAE Oxford ATPL says that 16,850ft is the answer.

The next closest answer is 15,750ft which is what I would choose.

Please help me out here, thank you in advance!

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Reference: CAE Oxford ATPL Chapter 8, Question 4.

  • $\begingroup$ FWIW I also calculated 18,691. I checked your speed calc using a E6-B, I checked your use of the chart I alos got 10.7% (ish) and I also go 18,691 ft. Quite often these questions in study material print the wrong answer $\endgroup$
    – Jamiec
    Jul 13, 2020 at 14:27
  • $\begingroup$ @Jamiec Thank you so much for your verification! Now I can sleep in peace. $\endgroup$
    – shogunnyan
    Jul 13, 2020 at 14:28

1 Answer 1


The relevant section (3.2.3) of CAP 698 explains how to calculate this as follows:

Distance to Reach given height.

To calculate the ground distance travelled in order to attain a given height above reference zero:
a) Convert the IAS 100 kt to a TAS, assume no position error.
b) Apply the wind component to the TAS to obtain the ground speed.
c) Determine the climb gradient from the graph.
d) Calculate the still air distance using the formula:
$$\text{Still Air Distance} (ft) = \frac{\text{Height Difference} (ft)}{\text{Gradient}} \times 100$$ e) Calculate ground distance using the formula: $$\text{Ground Distance} = \text{Still Air Distance} \times \frac{\text{Groundspeed}}{\text{TAS}}$$

Assuming your calculation of TAS is right, and your interpretation of the graph is right (which I agree with) you can plug this into the above calculations and confirm the answer is indeed:

Ground Distance = 15,995ft

However, if we play devil's advocate and reverse engineer the climb gradient from the expected answer, you'll see there might be some human/interpretation error at work.

Taking the published answer of 16,850ft and dividing by $\frac{\text{Groundspeed}}{\text{TAS}}$ gives us a ground distance of 19,690ft. Take this and Divide by 100 gives 196.9. Then doing $\frac{1}{196.9} \times \text{Height Difference} (ft)$ gives a resulting climb gradient of 10.2 - this is all just mathematics, no aviation knowledge required.

Now, can you say for sure that you've not misread/misused the chart? Once again I got the same result as you by looking at it in low def on a screen. But it's possible we both miscalculated the climb gradient.

Side note: I see question like this come up fairly regularly on other aviation forums, and a reasonable amount of time it is student error and a reasonable amount of time it is publication error. Make of that what you will.

  • $\begingroup$ Hmm interesting take on the answer. I will do the same and see what I come up with. Thank you once again @Jamiec. You sound like a ground school instructor hahaha. $\endgroup$
    – shogunnyan
    Jul 13, 2020 at 17:07
  • $\begingroup$ I'm still convinced that both our solutions are correct. There's no way the interpretation of the chart will give us 10.2%.I have been drawing the lines on the printed version and I have not got anywhere close to 10.2%. But anyway thank you for offering your knowledge, really appreciate it. $\endgroup$
    – shogunnyan
    Jul 13, 2020 at 17:31
  • $\begingroup$ I'm not even slightly a ground school instructor. It's been nearly 20 years since I had to learn this stuff, and embarrassingly I don't use it nearly enough as I tend to fly places where I know I have enough safety margin. But when I see a question like this I like to refresh my knowledge and try it out. $\endgroup$
    – Jamiec
    Jul 13, 2020 at 18:42

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