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Can somebody please explain how I go about this problem?

A Cessna 172S is flying at 9842.52 ft from the ground and the engines stop producing power.

If the aircraft is in a glide configuration and the best (quickest) rate of descent recommended is 3.4 m/s (800 ft/min) at a speed of 68 KIAS. What is the maximum range that can be reached?

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  • $\begingroup$ Related (dupe?): Is there a formula to calculate ground distance traveled given rate of climb and true airspeed? $\endgroup$ – ymb1 Mar 27 at 16:55
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    $\begingroup$ "and the engines" well, engine. Is this a no-wind situation as well? Head will reduce the range, tail will enhance, crosswind might do either depending on the direction. Assume no wind. 9842ft/800 ft/min = so many minutes of flying time. Minutes of flying time x 68 knots (1.15 miles/hours) = number of miles you can go. Do the math. $\endgroup$ – CrossRoads Mar 27 at 16:55
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    $\begingroup$ 9842.52 feet is bizarrely specific! Oh, wait, that's 3000 meters. $\endgroup$ – FreeMan Mar 27 at 17:16
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    $\begingroup$ You really want the quickest rate of descent? I'd expect that is a straight-vertical descent powered by gravity (aka, a nose-dive), with a total glide-distance of about Zero. $\endgroup$ – abelenky Mar 27 at 17:36
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    $\begingroup$ "Best rate of descent" is "slowest" not "quickest" $\endgroup$ – Adam Mar 27 at 19:16
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Basic Math:

How long will the plane be in the air?

$\frac{9842.52 ft\ \text{AGL}}{800\ fpm} = 12\ \text{minutes}$


How far will the plane go in 12 minutes?
$68kts \cdot \frac{12\ minutes}{60\ min/hr} = 13\ \text{NM}$

Assumptions

  • Assuming no tail wind or head wind, so that KIAS relates to actual ground-speed.
  • Approximating that 68 KIAS is the ground-speed, when it is in fact the slant-distance speed. If you need a precise answer, apply the Pythagorean Theorem to refine the answer.
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@abelenky's answer is correct if you're on the ground and therefore have time to look up figures and do the math. If you're in the air, the rule of thumb for light aircraft (with a typical glide ratio of 8:1) is 1.5nm per 1,000ft AGL, after subtracting 1,000ft or so for flying a pattern around your landing site. Rounding 9,842ft to 10,000ft, that gives a gliding range of 9×1.5=13.5nm, which is close enough in an emergency.

Note that the Vg (if any) given in your POH assumes max gross weight; if you're lighter, your best glide speed decreases and your gliding range increases. And of course, that all assumes no wind, which is unlikely in the real world; you'll generally want to glide downwind (to maximize range) and then do a 180 to land with a headwind, which is covered by the 1000ft subtracted above for the pattern.

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