I need answers to step 5 and 6 I'm completed questions 1 through 4.

Step 1:

Single Engine: 75 knots = 7595 fpm
v = (75 x 6076) / 60 = 7595 fpm
Airliner: 150 knots = 15190 fpm
v = (150 x 6076) / 60 = 15190 fpm

Step 2:

For the Sea Level Airport, all the temperatures are shown as 40°C, which would convert to 104°F.
For the Mountain Airport, 20°C = 68°F, 30°C = 86°F, 40°C = 104°F.

Step 3:

Sea Level Airport.

29.92 - 29.92 = 0
29.92 - 28.92 = 1 Hg. x 1000' = 1000 ft.
29.92 - 27.92 = 2 Hg. x 1000' = 2000 ft.
29.92 - 29.92 = 0
29.92 - 28.92 = 1 Hg. x 1000' = 1000 ft.
29.92 - 27.92 = 2 Hg. x 1000' = 2000 ft.

Mountain Airport.

29.92 - 23.92 = 6 Hg. x 1000' = 6000 ft.
29.92 - 23.92 = 6 Hg. x 1000' = 6000 ft.
29.92 - 22.92 = 7 Hg. x 1000' = 7000 ft.
29.92 - 22.92 = 7 Hg. x 1000' = 7000 ft.
29.92 - 21.92 = 8 Hg. x 1000' = 8000 ft.
29.92 - 23.92 = 6 Hg. x 1000' = 6000 ft.
29.92 - 23.92 = 6 Hg. x 1000' = 6000 ft.
29.92 - 23.92 = 6 Hg. x 1000' = 6000 ft.
29.92 - 22.92 = 7 Hg. x 1000' = 7000 ft.
29.92 - 22.92 = 7 Hg. x 1000' = 7000 ft.
29.92 - 21.92 = 8 Hg. x 1000' = 8000 ft.

Step 4:

68°F - 38°F = 30°F x 60 = 1800 ft. (then add to Airport Pressure Altitude for DA ROT answer).
86°F - 38°F = 48°F x 60 = 2880 ft. (then add to Airport Pressure Altitude for DA ROT answer).
104°F - 38°F = 66°F x 60 = 3960 ft. (then add to Airport Pressure Altitude for DA ROT answer).

Questions ~

Step 5:

How to linearly interpolate between two known values:

If we have the values of a parameter (for example rate of climb) at temperatures of 10 oC (where the rate of climb value is say 100 fpm) and 20 oC (where the rate of climb is 200 fpm), and we need to know the value of the rate of climb at 15 oC, we have to interpolate. We see that 15 oC is half way (50% of the way) from 10 oC to 20 oC. You use the following steps to interpolate:

Take the difference between the two known values (200 fpm-100 fpm = 100 fpm) and multiply that value times the percentage of 50% in decimal format (50% = 0.5) to get 100 fpmx 0.5 = 50.fpm.
Add the interpolation value in step (1) to the lower number (100 fpm) to get the final interpolated value; 100 fpm + 50 fpm = 150 fpm at 15 oC.

Note: The Take Off Roll data must be interpolated/extrapolated across pressures (vertically in the table) only; across-temperature interpolations (horizontally) will produce erroneous results. For the Rate of Climb data, interpolation/extrapolation is to be accomplished horizontally and vertically in the table.

Step 6:

The last calculation you will need to make is what altitude will your aircraft be at when it reached the obstacle in the path of flight after take off (T/O). In order to calculate the altitude your aircraft will be at when you travel the distance from takeoff to the obstacle, you need to use the following example as a model for making your calculations.

If you are traveling at 60 knots (nautical miles per hour), how far do you travel in one minute? We know that we travel a nautical mile (or 6,076 feet) a minute, which you get by dividing your speed (60 nautical miles per hour) by the number of minutes in an hour: that is 60 nautical miles per hour divided by 60 minutes per hour = 1nm per minute of 6076 feet per minute (fpm). Remember, you have already practiced this part of the problem to get speeds in fpm.

Knowing that you are traveling at 6,076 fpm, how long will it take you to travel 3 nautical miles (3 nm = 18,228 feet) from the point where you lift off of the runway (the end of your takeoff roll)? We know intuitively that it if we are traveling at 1 nm per minute it will take 3 minutes, but here is how you