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I have trouble understanding and calculating FPNM for standard take-off minimums.

As I understand, the formula is: (Ground speed)/60 * FPNM = FPM

Let's say the ODP says:

565' per NM to 2800, or 1000-3 with minimum climb of 370' per NM to 2800, or 1500-3 for climb in visual conditions

With a Cessna 172S, could you let me know whether the aircraft is able to meet this requirement? Here is the rate of climb chart:

enter image description here

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  • $\begingroup$ I suggest that you put some numbers into your formula and see if the result makes sense. If not, or if you don't understand the result, feel free to to add it to your question: a specific example would help us to understand where your problem is. $\endgroup$
    – Pondlife
    Nov 21, 2018 at 5:45

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If your takeoff speed is 60 knots, and you're climbing at 500 feet per minute, then you are climbing at 500 feet per nautical mile, yes? Extrapolate from there.

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    $\begingroup$ Not exactly. Feet per nautical mile is based on ground speed, not airspeed. And like the chart shows, fpnm varies with temperature and altitude. $\endgroup$
    – JScarry
    Nov 22, 2018 at 3:42
  • $\begingroup$ @JScarry, Crossroads used "knots", not IAS. (knots could also mean groundspeed and airspeed could be in MPH...) Granted GS in knots wasn't specified, but neither were winds. Given the simplicity clearly intended to make a point, it's pretty safe to assume a no-wind situation was meant here. $\endgroup$
    – Michael Hall
    May 19 at 16:18
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You don’t specify an altitude in your question, so let’s start with 2000' and the requirement to climb at 565' per NM and assume you have the airspeed at 73 kts. Then the performance table says

-20°   0°   20°    40°

760   695   555   560

So you meet the requirement for all temperatures up to almost 40°.

At 4000' you will make it at 20° but to see what the maximum temperature would be you need to interpolate.

-20°   0°   20°   ?°   40°

685   620   625  565  496

There are lots of ways to do the arithmetic, one way is with percentages. The difference between 625 and 496 is 129. The difference between 625 and 565 is 60. So 565 is 60/129 (47%) of the way between the two. Our unknown temperature is also 47% of the way between 20° and 40°.

The difference in temperature is 20°. 47% of that difference is 9.4°. We went from left to right in calculating the percent difference so we do the same thing for temperature and add 9.4° to 20° to get 29.4°.

So as long as the temperature is less than 29.4° and you maintain an airspeed of 73 kts indicated, you can meet the climb gradient.

If you are at an altitude in between one on the table then you would need to do two interpolations. One at the lower altitude and one at the higher.

In the real world, and not on an FAA test, just use the next higher altitude and temperature.

Note that the table is based on pressure altitude, so set your altimeter to 29.92 to get the pressure altitude of the field.

Also note that if your airspeed is faster than in the table, you will be climbing too slow to make the numbers. This is also a no wind table. If you are climbing with a tail wind you will not meet the climb rate per nautical mile.

And finally, the climb can be to an altitude that is fairly high above the field elevation. If you look at the chart you will see that performance drops off with altitude. In your example if the field is sea level, then you will need to climb 2800' at 565 fpnm. Don’t use the sea level numbers since you are climbing through the 2000' level. Use the 2000' numbers to be safe.

My home field is at 212' MSL and one of the departures has a minimum climb of 320' per NM to 3000'. I’d use the 4000' row to decide whether I could make the departure.

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  • $\begingroup$ The response from JScarry above is good, but I think the Cessna table provides values in Feet/Minute, so you still need to convert that FPM to Feet/Nautical Mile. Be careful. $\endgroup$
    – Brian
    Dec 20, 2022 at 15:53
  • $\begingroup$ Downvoted for confusing FPM and FPNM. Fix and I’ll reverse because this is a good discussion of interpolation. $\endgroup$
    – StephenS
    Dec 21, 2022 at 15:12

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