Hi there!
I was going through my CPL nav KDR and have a few questions.
The first is ETA.
You are at A at Time 1200, flying to B which is a distance of 25 nm, planning to arrive at B at time 1215 however, you are 2 min Late at B. B to C 50 Nm. Recalculate ETA C.
I thought this was proportional, making time at C, 2 x 2= 4 + the original 2 mins late = 6 mins late, time 1221. What is the correct answer, as I think I got this question wrong.
Thank you so much!
Your approach assumed that since B to C is twice the distance of A to B, the delay doubles. However, the delay only increases if speed is reduced. Since speed remains the same, the delay is simply carried forward.
Work out the ground speed A-B
25/17 = 1.4705882353 nm/min
Now 50nm at 1.4705882353 nm/min = 33.9 so 34 Minutes
ETA is B @1217+34 = 1251 @ C
Gross error check total 75nm @ 1.4705882353 nm/min = 51 so 1200+51= 1251
Hey there,
Hope you are well.
Did you use a whizz wheel for this? The answer you have provided was not a choice on the multiple choice exam. From memory, the answers were something like 1216, 1217, 1218, 1219, 1220 and 1221. Thanks!
It always pays to draw a picture
Plan
A 25 nm B 50 nm C
Plan ETA 1200 1215 (twice the distance so twice ETI 1245
ATA 1200 1217 so first 25 nm took 17 mins, so next 50 nm will take double i.e. 34 mins.
thus revised ETA C is 1217 plus 34 = 1251
There can be a bit of trickery in theses questions, was it the revised ETA C or ATA B. A play on words.
Cheers W
Hey Wayne,
This is great and I understand the theory. However it was more along the lines of ' if you are 2 mins late to b, how many mins late will you be to c' given the distance from b to C is double that of a to b and you were to minutes late at B.
Is this something someone can help me with?
Thanks!
