Hey all,

Am having a bit of trouble interpolating. Excercise 4.3, Question 2, Page 4.6:

Find the climb gradient as a percentage for a maximum rate climb in an ECHO under the following conditions in nil wind:

Altitude= 5000
QNH= 1013
OAT= +25 degrees celsius
GROSS WEIGHT= 2725kg

I've seen the answers for the RoC and TAS, is there anychance you can show me how you worked those out? I get close but not the same answer, I'll try and explain what I do by adding an attatchment, and incase that doesn't work I typed out what I do step by step: (I tried making a table to make it a little more neat but it didn't let me)

Step 1: 1400-1500= -100
-100/5000= -0.02
-0.02x2400= -48
-48+1500= 1452

Step 2: 1950-2100= -150
-150/5000= -0.03
-0.03x2400= -72
-72+2100= 2028

Step 3: 2028-1452= 576
576/450= 1.28
1.28x225= 288
288+1452= 1740

Answer= 1740fpm


I used the same method to find the TAS.

Am sorry if that seems extremely confusing I got taught something like this when I had to interpolate for take off and landing distances when using the POH. Just want to know if it's correct/incorrect

Also for question 5 of the same excercise I got a gradient of 19.95%. Not too sure what I'm doing wrong I used the same way as I did for question 2.


Any help is always appreciated

Thanks

Firstly, don't get too excited about interpolation. In the CASA exam they don't give you the climb or TAS tables so rate of climb and TAS will simply be given in the question stem. I included it because the syllabus says you should be able to extract information from performance data.

Regarding the question, take a practical approach. Pressure height 5000 feet and ISA +20 gives a density height of 7400 feet. So interpolate between the 5000 feet figures and the 10000 feet figures. That will give you a rate of climb and ground speed to use to calculate the climb gradient resulting.