Interpreting Climb Data Chart

[AggieMike88]

For an older 172-I, I am presented with the following climb data chart.

I need to compute my Top of Climb info from 642-MSL elevation to 3,500 feet.

I know how to compute time and distance traveled based on winds, temps, pressures, etc. But interpreting the column "Gallons of Fuel Used" has got me a bit flummoxed.

If we use Sea Level and Max weight, is the chart saying that for every 645 feet I've climbed, I will have burned 1.0 gallon of fuel?

And if that assumption is correct, and using some rounded numbers Aggie Math (and leaving out wind and pressure altitude items), climbing from 500 feet to 3500 feet means it will take 4.65 minutes (3000 ÷ 645). And I'll burn 4.65 gallons of fuel getting up there. Correct?

 

[dmspilot]

Gallons of fuel used is the quantity of fuel needed to climb from sea level to the altitude shown at the top of the column. For example, it takes 11.5 gallons to climb from sea level to 15,000 feet at 2300 lb. Why does it take a gallon of fuel to climb from sea level to sea level, you might wonder? See note 2.

 

[AggieMike88]

Gallons of fuel used is the quantity of fuel needed to climb from sea level to the altitude shown at the top of the column. For example, it takes 11.5 gallons to climb from sea level to 15,000 feet at 2300 lb. Why does it take a gallon of fuel to climb from sea level to sea level, you might wonder? See note 2.

So I interpolate left to right?



And for Sea Level to 3,500 MSL, Fuel Used would be 1.82?
3500÷5000=0.7
0.7x2.6gal=1.82gal

 

[dmspilot]

So I interpolate left to right?

Yes.

And for Sea Level to 3,500 MSL, Fuel Used would be 1.82?
3500÷5000=0.7
0.7x2.6gal=1.82gal

Since a climb to sea level is 1 gallon and a climb to 5,000 is 2.6 gallons, the interpolation should probably be:
0.7*(2.6 gal - 1.0 gal) + 1.0 gal = 2.12 gal.

 

[Clip4]

The table starts at 1 gallon because the data includes warm up and takeoff. That value is not reduced for climb to a value between two knowns. You also have to determine fuel from SL to 645 and subtract that from fuel from SL to 3500.

1. Identify the two field to be used. (SL and 5000)
2. Subtract 0 (sea level) from 645= 645
3. Subtract 0 from 5000 = 5000
4. 642/5000= .1284 (interpolation)
5. 1.6 gallons x .1284= .2. (The fuel we don’t need to get to 645 ft)
6. 3500/5000= .7 (another interpolation)
5. 1.6x.7=1.12 (the fuel we need to get from sea level to 3500 ft
7. 1.12-.2= .92. ( fuel required from 645 to 3500) + 1 (takeoff warm up) = 1.92

 

[AggieMike88]

@Clip4 and @dmspilot .....

Thanks!

 

[Clip4]

@Clip4 and @dmspilot .....

Thanks!

Except we didn’t answer your problem because we solved from SL to 3500.. My edit solves from 645 to 3500.

 

[RussR]

Except we didn’t answer your problem because we solved from SL to 3500.. My edit solves from 645 to 3500.

Minor point - he is departing from 642 MSL, not 645. 645 is the rate of climb on the chart. Just so everybody understands which numbers are used for the calculations.

 

[Clip4]

Minor point - he is departing from 642 MSL, not 645. 645 is the rate of climb on the chart. Just so everybody understands which numbers are used for the calculations.

Math without my readers at midnight, misread the airport elevation he stated, value corrected.