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Is it possible to calculate the ground roll and distance to climb to 1000 ft in a Cessna 172S?

Why? I live near a small airport with 2 active flight schools. As part of a new noise abatement procedure the circuit path will be moving from 500 ft altitude to 1000 ft altitude with no turns below 1000 ft weather permitting. I'd like to be able to calculate some typical scenarios that will result in determining approximately where the end of the departure leg and the start of the crosswind leg of the circuit will happen to understand what the net change will be. If I can get the ground distances, I can use the scale line drawing tools on a Google Earth satellite image to draw in some eastward and westward ranges for where the crosswind leg might start, for example on a typical sunny early September morning like today.

I'm not a pilot. I'm a retired computer systems techno-geek who doesn't mind plunking parameters into formulae to crank out numbers. I'm just having trouble finding the formula or reference chart.

I've downloaded a Cessna 172S Pilot handbook and have the performance info. I can get current weather conditions for the airport thru their AWOS or from aviationweather.gov. I have a binder half full of reference material, sample pilot exam questions, definitions and terminology. I have the current aerodrome chart for the airport. I'm starting to muddle my way thru calculating the ground roll and distance to clear a 50 ft obstacle using that reference material.

What I can't find is anything that extrapolates that and figures out the distance along the ground to climb to what is essentially a 1000 ft obstacle.

Can anyone help?

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  • $\begingroup$ From what I understand, what you want can be calculated using your ground speed and your climb rate (something around 700 fpm for a 172). But be aware that your climb rate will vary depending on a bunch of factors like temperature and pressure altitude. The same is true for your takeoff distance. It will change depending on airplane weight, engine performance, etc. $\endgroup$
    – Jimy
    Commented Sep 1, 2017 at 18:06

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The POH published ground roll is 960 feet. (Performance Specifications, cover page)

The rate of climb from Sea Level is 730 Feet Per Minute, which is 82 seconds to climb to 1000 feet. (Section 5-18, Time, Fuel and Distance to Climb)

During that 82 seconds, the plane will have a airspeed of 74 kts, or 85 MPH, or 1.419 miles per minute.

So in 82 seconds, covering 1.419 miles per minute, the plane will cover 1.93 miles.

1.93 miles of climb, + 960 feet of ground roll = 11,200 feet, or 2.12 miles.

The last column of the table in Section 5-18 is "Distance in NM from Sea Level", which says that the distance it takes to climb from Sea Level to 1000 feet is 2 NM.

2 NM is 2.3 statute miles, which is pretty darn close to my calculation of 2.12 miles (difference of 950 feet). So the answer cross-checks nicely.

From start of takeoff roll to 1000 feet MSL, you can expect about 2.1 - 2.3 miles, not counting headwind and temperature factors.

I re did the calculations, accounting for a 20 kt headwind. (I think that is the maximum that most Cessna students would take off in), and that shortened the distance to 1.5 miles.

With max headwind: 1.5 miles
With no wind: 2.3 miles.

I think that covers your range pretty well.

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  • $\begingroup$ If you are not near Sea Level, you can use a different row from the same table. If you want, you can interpolate, or even integrate the climb between two different altitudes, but that would change the answer very little. $\endgroup$
    – abelenky
    Commented Sep 1, 2017 at 17:44
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    $\begingroup$ This is a good worst-case estimate. Since the airplane is climbing, it is following the path of the hypotenuse of a right triangle. The ground track will be slightly less. The ground track will also be less if there is any headwind. $\endgroup$
    – JScarry
    Commented Sep 1, 2017 at 17:49
  • $\begingroup$ I believe the result would be more accurate using the airplane's ground speed instead of airspeed. Since I assume the OP is about the distance on the ground (but I might be wrong). Great answer anyways! $\endgroup$
    – Jimy
    Commented Sep 1, 2017 at 18:04
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    $\begingroup$ @JimyPP, yes, but I have no way of knowing the ground speed since I don't know the headwind. My calculation shows the factors that go into it, and the poster can incorporate headwind himself. $\endgroup$
    – abelenky
    Commented Sep 1, 2017 at 18:07
  • $\begingroup$ Thank you everyone. If I were to incorporate temp & wind speed (& pressure?) into the calculation, is there a formula for that? I'd like to get at least 4 plot points: 30C, no wind; 30C, wind 11 knots NW 310 degrees; 10C, no wind; 10C, wind 11 knots NW 310 degrees. That would help establish a reasonable range for the end of the departure leg accounting for climate. Is it worth doing all those calculations or could I just I could use 2.12 miles + or - a percentage to account for temp and wind. Any ideas on what that + or - percent might be? $\endgroup$
    – Kinmount
    Commented Sep 1, 2017 at 20:46

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