The 15th December is the last day to purchase Extra Access exams, physical books and combo deals until the 8th January 2024
as the office is closed for this period

×
Welcome to the CPL Performance question and answer forum. Please feel free to post your questions but more importantly also suggest answers for your forum colleagues. Bob himself or one of the other tutors will get to your question as soon as we can.

Just wondering with this question in the exam (Exam 2 Q13) I havent really been able to find too much on this sort of stuff in terms of figuring out if you have enough room to clear objects as such.

The answer in this question talks about 5% as a (1 in 20 Gradient) Fair enough, But was if its elsewhere from 5% like 3, 4 or 6 I cant really figure out what comes after that

I'm not entirely sure that I understand the problem, however, let's start with the following and take it from there.

It may help if you redraw the graphic with a triangle (the two existing dimensions are the base and rise, while the long side is the slope which the aircraft has to stay above. That gives you a better idea of what you are trying to do with the aircraft.

The 5% thing is a very old airport standard and fits in with the (now defunct) requirement for a light aircraft to have a 6% minimum climb gradient capability (all engines operating).

In the question case, the gradient is 9 x 100 / 180 = 5 %. The 100 factor just converts the fraction to a percentage.

If you are looking at some other gradient, you just change the numbers to suit. So, for example, if you wanted to consider, say, a 4% gradient, you might have a rise of 8 metres and a base (or run) of 200 metres. In this case, 8 x 100 / 200 = 4 %.

If you are looking at a paddock takeoff operation (which is the original consideration for this sort of stuff), then you have to ensure that you juggle the numbers to achieve the requirement. This should be 5 % to meet the airport requirements. In practice, you adjust the distance so that the run left from the end of the TODA to the obstacle's rise height will result in the required gradient.

Describing the gradient as 1:20 rather than 5% is just another way of saying the same thing. For this, we don't want the answer as a percentage so we leave out the x 100 factor in the sums.

So, 9 x 100 / 180 = 5 % would become 9 / 180 = 1 / 20. Just like changing from clear glasses to sunglasses when you are looking at something ... same, same.

If that discussion didn't help, then perhaps you can restate the question in some other way and we can try again ?

Engineering specialist in aircraft performance and weight control.

This is awesome thanks John. Great help thank you. I've had ago at trying to apply this knowledge else where and I think I may have got it right. I ended up creating a triangle around it to help with the understanding. Let me know if this isn't correct in anyway to do.

Let me know what you think of this working out as well

However, the question is ill-considered and a bit silly. It appears that the writer of the question doesn't quite understand the requirements ?

The ground run, basically, is irrelevant when considering small aircraft; one is interested in the TODR and, although we might infer some numbers from the old 6% WAT requirement, that's a bit rubbery for the question.

However, clearly the available distance is inadequate. If we don't have enough to account for the ground roll, then the TODR becomes a bit problematic.

Engineering specialist in aircraft performance and weight control.