The 172M POH doesn't have a "Distance to climb" chart like the newer 172s. I just have the climb rate (Call it 600 ft/min) and so I estimated a time of 8.5 minutes. The problem is, even if I calculated the distance (about 1 NM), it's the vertical distance which I know, not the ground distance which I'm looking for. Do I have to use trigonometry to calculate my ground distance?

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    $\begingroup$ You don't need to use trig if you have time and speed... Think of it this way, if you were NOT climbing, how much distance would you cover in 8.5 minutes at __ airspeed? $\endgroup$ May 3, 2022 at 4:21
  • $\begingroup$ Related: Is there a formula to calculate ground distance traveled given rate of climb and true airspeed? $\endgroup$
    – user14897
    May 3, 2022 at 9:43
  • $\begingroup$ How precise do you need? $\endgroup$ May 3, 2022 at 23:23
  • $\begingroup$ Actually, distance to climb involves 2 critical components of preflight planning, weight and weather. Climb rate and distance to climb are heavily influenced by these factors. "Simplifying" things does not equal skipping important steps, nor does it mean taking a little time to understand why a little "shorthand" works for a 172. Distance to climb becomes important if terrain or something like a radio tower (windmills these days) is in the way. $\endgroup$ May 4, 2022 at 11:40

4 Answers 4


One key point that has not been explicitly mentioned so far, though it's apparent from the example given in another answer, is that for light piston-engine airplanes, the angle of climb is so shallow that there's no need to go through the step of using trigonometry to calculate your climb angle to convert the speed and distance travelled along the flight path to the speed and distance travelled over the ground. For example if you are climbing at 70 knots true airspeed, just assume that the horizontal component of the airspeed vector is also 70 knots. Then you are ready to factor in the wind to calculate your groundspeed, and your distance travelled over the ground.

Here are some numbers to help make this point clearer-- in still air, to make the horizontal component of your airspeed vector be 5% smaller than your total airspeed vector, you'd need a climb angle of 18.2 degrees. At 70 knots airspeed, this would require a climb rate of 22.9 knots, or 2210 feet per minute-- far beyond the capacity of a Cessna 172.

  • $\begingroup$ Your opening sentence mirrors my comment. No need for trig on a light single, keep things simple. You got my upvote. $\endgroup$ May 3, 2022 at 14:37
  • $\begingroup$ (re "capacity" -- "capability" might be a better word) $\endgroup$ May 3, 2022 at 17:27

You would typically use four pieces of information to calculate the distance to climb:

  • start altitude
  • altitude to reach
  • vertical speed
  • ground speed**

First you calculate the total altitude to gain:

  • altitude to gain = altitude to reach - start altitude

From the altitude to gain and the vertical speed, you can calculate the time to reach the target altitude

  • time to climb = altitude to gain / vertical speed

Now that we know the time it will take to climb, we can calculate the horizontal distance we will travel during the climb:

  • horizontal distance = time to climb * ground speed.


  • ground speed = true air speed*cos(climb angle) +- along track wind;

Note that the cosine of a small angle is approximately 1. Since the climb angle is shallow, we can ignore the effect of the climb on the forward speed component. For example if the climb angle is 10 degrees, the forward component of the speed is cos(10°) = 0.985 ≈ 1. We can therefore leave out the trigonometry and work with the simplified:

  • ground speed = true air speed +- along track wind;


  • start altitude = 1000 ft
  • altitude to reach = 4000 ft
  • vertical speed = 600 fpm
  • ground speed = 120 kts

Using the formulas above:

  • altitude to gain = 4000 - 1000 = 3000 ft
  • time to climb = 3000 / 600 = 5 minutes
  • horizontal distance = 120 * 5/60 = 10 NM

Note that this only works for relatively short climbs, otherwise effects of wind and air density will change both the ground speed and the climb performance during the climb. To estimate the climb distance for longer climbs, you can cut it into multiple segments of shorter climbs, and use appropriate vertical speeds and ground speed for each segments. However, this is not an quick exercise that is easily done in the cockpit.

  • $\begingroup$ Other answer have mentioned the key point that ground speed ~= true air speed +- wind, because climb angles are so shallow. But your answer seems to omit it. I'll suggest an edit to make this answer self-contained. $\endgroup$ May 4, 2022 at 2:15
  • $\begingroup$ @PeterCordes, thank you for the suggested edit. $\endgroup$
    – DeltaLima
    May 4, 2022 at 8:59

Well, you have your altimeter and your watch, so rate of climb can be exactly determined.

Distance covered on the ground is the $cosine$ of distance in the air.

Here you're in luck as the 172 has an angle of climb of around 6 degrees.

This is calculated as arc sin of 600 fpm climb/6000 feet per minute airspeed.

The cosine of 6 degrees is 99.5%, which means your ground distance is 99.5% of what you flew in the air.

The slightly trickier part is that true airspeed, not indicated, must be used for more precise measurements.

In a 172, it might be easier to look out the window for a landmark or use GPS for distance traveled. This may be advisable considering head or tail winds will also skew any distance calculations.

  • $\begingroup$ Oh right, even though 1 part in 10 (climb speed as a fraction of airspeed) isn't tiny, 0.1^2 + x^2 = 1.0 is still only a tiny impact on the x component (for coordinates on a unit circle, which is why cos^2 + sin^2 = 1.0). So yeah, cos(climb angle) = 0.99499 = sqrt(1 - 0.1^2) $\endgroup$ May 4, 2022 at 2:26

While it is good to understand all the nuances involved in these sorts of questions, rule #1 when flying an airplane is to keep things simple so that you can maintain situational awareness of the big picture. Approximations and rules of thumb can, and should be used liberally.

That said, at the climb angles of a typical light single GA airplane, small angle approximation lets us dispense with Trig and turn this into a simple rate problem: If you are flying at 60 Knots for the 8.5 minutes in your example you will have covered 8.5 nautical miles. Double the speed to 120 and you cover twice that distance, or 17 nautical miles. (don't forget to add/subtract wind for distance over the ground!)

Is that too fast or slow for your 172? No problem; like Goldilocks pick an airspeed that’s just right such as 90 knots and split the difference for 12.75. (Then round up to 13. Or, easier mental math might be to round 8.5 down to 8, 1.5 of which is 12, then add a mile for some cushion to cover the 30 seconds you chopped off...)

SUMMARY: You should be able to rapidly calculate in your head approximately how much distance you will cover in any given span of time. (whether in climb, cruise, or descent) For typical small general aviation aircraft the easy benchmarks to memorize are –

  • 60 = 1 mile per minute
  • 90 = 1.5 miles per minute
  • 120 = 2 miles per minute

Knowing the above keeps the mental math simple, and allows you to approximate values in between.

It isn’t just useful for climbs, in fact estimating your Top of Descent (TOD) is just as important. For example, let’s say you are cruising at 120 knots at 4000’, and planning a 500 FPM rate of descend to an airport at sea level. It will take you 8 minutes to complete the descent, so you would want to start down around 16 miles out.

While trigonometry is very useful, I would strongly recommend against performing complex calculations in flight unless you are very comfortable with it. And keep in mind that the combined effects of varying winds, TAS vs IAS, plus your ability to hold airspeed within +/- 5 KIAS will likely have a greater effect on actual vs calculated distance than the difference between the adjacent leg and the hypotenuse.


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