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The descent rate calculation states that descent rate is calculated by 'groundspeed / 2 * 10'.

Another document stated that the ground speed should be multiplied by 5 (or more accurately 5.2) to acquire the descent rate. The 5.2 value is found by finding the gradient for 3 NM for 1000 feet (based on the 3:1 glide ratio), which is 5.2% (found using tan of 3 degrees). The document can be found here: https://www.ivao.aero/training/documentation/books/SPP_APC_Top_of_descent.pdf

My question is: How was the formula created? It's interesting that I can't find any info on this on the internet. Can someone show the maths behind this formula?

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    $\begingroup$ groundspeed / 2 * 10 is the same thing as groundspeed * 5, which is probably a "close enough" approximation for a pilot to do in their head rather than using the 5.2 value. $\endgroup$
    – Ron Beyer
    Commented Sep 9, 2016 at 3:17
  • $\begingroup$ I was aware that it was equal to groundspeed * 5, but I included that as a because that is what most people are familiar with. I know that it is the gradient calculated by tan(3°), but I would like to know the maths behind it. Thanks for your comment. $\endgroup$
    – user16863
    Commented Sep 9, 2016 at 5:44
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    $\begingroup$ @WeavingBird1917 would I be right in thinking your also after information on where the 3 degrees came from? $\endgroup$
    – Notts90
    Commented Sep 9, 2016 at 8:31
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    $\begingroup$ @Robert If you are referring to why 3 degrees is the optimal glide ratio, then yes, that would certainly be interesting! $\endgroup$
    – user16863
    Commented Sep 9, 2016 at 9:47
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    $\begingroup$ @WeavingBird1917 I thought so too so just asked it. aviation.stackexchange.com/q/31382/15982 $\endgroup$
    – Notts90
    Commented Sep 9, 2016 at 10:13

4 Answers 4

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From Gypaets' answer

$$DescentRate = 0.0524 \cdot GroundSpeed$$

But the Ground speed is in nautical miles per hour [NM/h], and the descent rate is in ft/min. So we have

$$DescentRate \left[\frac{ft}{min}\right] = 0.0524 \cdot GroundSpeed\left[\frac{NM}{h}\right] $$

With $6076.12 \left[ft\right]$ in 1 $[NM]$ (yay for imperial units) and $60$ $[minutes]$ in 1 $[h]$ we get a conversion factor: $$ 1 \left[\frac{NM}{h}\right]= \left[\frac{6076.12~ft}{60~min}\right] = \frac{6076.12}{60}\left[\frac{ft}{min}\right] \approx 100\left[\frac{ft}{min}\right]$$ This leaves: $$DescentRate \left[\frac{ft}{min}\right] \approx 0.0524 \cdot GroundSpeed \left[\frac{NM}{h}\right] \cdot 100_{(NM/h~to~ft/min)} $$

Further approximating it to:

$$DescentRate \left[\frac{ft}{min}\right] \approx 5 \cdot GroundSpeed \left[\frac{NM}{h}\right] $$

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  • $\begingroup$ Yeah I was troubled a bit expressing the units and the conversions. I tried to make it clearer now $\endgroup$
    – ROIMaison
    Commented Sep 9, 2016 at 9:13
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    $\begingroup$ That was what I was looking for! That explains why the 'tan value' is multiplied by 100. The document I linked in my original post converted the value to a percentage, but it never stated that it was to approximate the conversion of NM to ft. $\endgroup$
    – user16863
    Commented Sep 9, 2016 at 9:55
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Just take a look at the geometry, the ground speed and descent rate vectors are perpendicular to each other. enter image description here

If $\gamma$ is the descent angle, the formula you are interested in is:

$$tan(\gamma) = \frac{DescentRate}{GroundSpeed}$$

For $\gamma = 3°$

$$DescentRate = 0.0524 \cdot GroundSpeed$$

Obviously, you have to convert the result to the desired units.

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  • $\begingroup$ Thanks, could you explain in a little bit more? For example, if my TAS was 250 kt, multiplying it with tan of 3 degrees would give me 13.1. I am interested in how the descent rate formula came about, and I am also unsure how it manages to give less than a 5% error, especially with the differences in the unit of measurements (nm and feet). $\endgroup$
    – user16863
    Commented Sep 9, 2016 at 7:30
  • $\begingroup$ @WeavingBird1917 I added a picture so you can see the $tan$ clearer. $\endgroup$
    – Gypaets
    Commented Sep 9, 2016 at 9:51
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    $\begingroup$ Thanks, that made it much clearer! I marked ROIMaison's post as the answer because it covers the conversion and why the tan value is multiplied by 100. $\endgroup$
    – user16863
    Commented Sep 9, 2016 at 10:04
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So with 3 degree,

500 ft/min descent rate = .0524 * Gnd speed

500/.0524 = 9541 ft/min * 60 min/hr * 1 mile/5280 ft = 108 mile/hr Gnd speed

Work it the other way to find descent rate given speed

Using 250 mph (as I don't have the knots-mph conversion rate handy)

descent rate = .0524 * 250 miles/hr * 1/60 hr/minute * 5280 ft/mile = 1153 ft/min

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radian rule is theta=l/r where theta is radian

now theta= ROD(ft per min)/GS(kts)

theta/60 X GS (Kts) = ROD (ft per min) (here theta is degree as it is divided by 60)

theta/60 * GS (Kts) * 6080/60 = ROD (ft per min) (6080 for NM to Feet) (6 for hr to min)

now replace theta with 3 GS (Kts) * 5 = ROD (ft per min)

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