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How do I calculate the distance to touchdown when the glideslope angle is different from 3°? For example, you are 670ft AGL on a 3.2° glideslope approach; what do I change in the formula to calculate the distance?

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    $\begingroup$ Which formula are you referring to? $\endgroup$ – abelenky Jan 15 '14 at 15:51
  • $\begingroup$ How about doing it for 3° then subtracting 7% (a close approximation of .2)? $\endgroup$ – Simon Jan 15 '14 at 16:20
  • $\begingroup$ @Simon subtracting 7% is quite fair (see also my answer on approximation). $\endgroup$ – yankeekilo Jan 15 '14 at 18:40
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Just in case you don't have a calculator at hand (yeah, I know...):

WolframAlpha returns (unsurprisingly): 1.97nm (and lot of other entertaining info).

Mathematica graphics

Sidenote: Because $\tan(\alpha)\approx\alpha~for~small~\alpha$, you can use a simple linear approximation in this case. Your new glide distance will be more or less $\frac{3}{3.2}=\frac{15}{16}=0.9375$ times the original one.

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    $\begingroup$ Or paste this into google. 670 ft/tan(3.2 degrees) to nautical miles $\endgroup$ – Magnetoz Jan 15 '14 at 20:26
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The $300ft/NM$ used to calculate your height above the runway at a particular distance is just an approximation that is easy to remember, and is close enough for our purposes.

If you want a more exact number, you can use some trigonometry:

$$\tan(3)=\frac{xft}{6,076ft/NM}=318ft/NM$$

(the "real" number).

For 3.2 degrees, it would be:

$$\tan(3.2)=\frac{xft}{6,076 ft/NM}=340ft/NM$$

So if you want to know the appropriate altitude for, say a 3 mile final on a 3.2 degree glideslope, it would be:

$$Runway~Touchdown~Elevation+340ft/NM\times3NM=TDE+1,020ft$$

In your specific example where you already know the altitude and want to know when to start the descent, the answer would be $2NM$:

$$\frac{670ft}{340ft/NM}=2.0NM$$

($1.97$ if you want to be REALLY exact.)

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Since the 1 in 60 rule of thumb yields a solution of 300 feet for every NM from the touchdown point for a 3 degree slope, use 320 feet for a 3.2 degree slope. So 670ft, gives you a nadgers whisker over 2 nm.

To be a little more precise, 2.09375 nm ($\frac{670}{320}$)

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    $\begingroup$ DeltaWhisky, not sure why your edit disappeared but I must point out that a nadgers whisker has a very precise meaning in general aviation. A badgers whisker is quite different, so I've edited it back. $\endgroup$ – Simon Jan 15 '14 at 16:47
  • $\begingroup$ Just out of curiosity, how did you determine that a 3.2 degree slope needs a 320 ft/NM descent? Is there a rule of thumb that you are using to determine that? $\endgroup$ – Lnafziger Jan 15 '14 at 22:02
  • $\begingroup$ As you say. it's just an approximation but 300ft per NM at 3 degrees is good enough for GA. It's an extension of the 1 in 60 rule. So, you can say 100ft per degree. 3.2 degrees is therefore 320 ft. $\endgroup$ – Simon Jan 17 '14 at 12:31
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$$tan(3.2)=\frac{670}{a}$$ $$\frac{670}{\tan(3.2)}=a$$

$$a=11,983ft$$ $$11,983ft=1.97NM$$

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Since we all agree that a 3.2°GS give us 340'/nm, I suggest that you take your FAA, subtract by the TCH elevatin and divide by 340, the result is the distance to touchdown

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Thanks to all for answers. The $\frac{1.97}{96}$ was not correct.

This is how you should do it:

$$3.2=\frac{670ft\times60}{3.2}=12,562ft=2.066NM$$

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    $\begingroup$ The 1.97 nm is the precise value. The rule of thumb overestimates the value slightly. $\endgroup$ – Jan Hudec Mar 12 '14 at 19:19
  • $\begingroup$ As Jan says, 1.97 is the actual correct mathematical value. Your answer does not accurately answer the question as you originally asked it.... $\endgroup$ – Lnafziger Mar 12 '14 at 19:25

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