Quick way to calculate ROD

[Founder]

Hi, does anyone have quick and easy way to calculate ROD if you know the
descent angel or gradient and the GS...

for example:
Gradient 12%
GS 540 kt

are there any good formulas to memorize..?

/Tim

[Pittsle]

For me the following works pretty good in my head (though mathematically probably not exactly correct):

Based on the nautical miles per minute e.g. 540 kt equals 9 nm/min (or mach number at higher levels).
1 degree change of flightpath times 9 equals 900 ft/min descent /climb rate.

To calculate the flightpath-angle: Altitude to loose in FL (e.g. 20000ft 200FL)
divided by distance to go e.g. 50nm equals 4 degrees.

So if you want to loose 20000ft in 50nm at a speed of 540kt (or .9mach) you need 4 degrees nose down which gives 3600 ft/min descent rate.

Cheers

[Lemper]

Old stuff from the days we flew with butts and attitude:

At high speed (lo AOA) 1° pitch down = 1000 fpm. Try.

At lo speed (appraoch), there is a table at the bottom of the Jepp 11-* pages

[florida flamingo]

GS(Kts)xgradient(%)=ROD(Ft/min). Just remember that for a constant IAS decent the TAS will be reduced when altitude decrearses and thus for a constant gradient the ROD will be reduced.

[super 27]

there's another technique if you're using a 3% gradient just do GS/2times10 gives ROD.For example GS 540kts/2=270 times 10=2700 ft/min

[Blip]

super 27 do you mean to say a 3 degree slope not 3 percent?

A 3 degree slope = 1:20 = 5%

540 kt x 10/2 = 2700 fpm.

540 kt x 5 = 2700 fpm.

Therefore what florida flamingo said! :)

[willnotcomply]

Super 27 is on the money with his quick calculation. It works in a C152 or 744. You other guys are making your lives difficult.

[212man]

Will, RTFQ: Super is mixing his percents with his degrees!

Ballpark figure 1 degree = 1.7%, so 3 deg = 5%.

multiply gradient in % times G/S in kts to get RoD (so yes, for 3 degrees, that equals GS*10/2)

[SFI145]

It is very much simpler than all the previous posts.

Gradient in % x G/S kts = Required rate of descent or climb in ft/min

This is why 5 x G/S is the correct rate of descent on an ILS (a 3 degree glideslope is approximately 5%)

Sorry to simplify things as I know PPruners love to make everything esoteric.

[willnotcomply]

SFI145 Well said!

[212man]

It is very much simpler than all the previous posts.

Gradient in % x G/S kts = Required rate of descent or climb in ft/min

Ah, yes that's entirely different to what I said:multiply gradient in % times G/S in kts to get RoD:ugh:

[757_Driver]

for those that want to know why you can multiply knots (nautical miles per hour) by the gradient and come up with feet per minute. its because there are give or take 6000 ft in a nm and there are 60 minutes in an hour.
The six's cancel out. so you can just to %grad x ground speed.
i,e 140 kt approach times 5% gradient = 700 fpm

FWIW
the genuine answer for a 3 degree approach (5.2%) is 140*5.2% = 7.28 nm / hr * 6080 ft = 44262 ft / hr. divided by 60 mins per hour gives 737 ft /min
I think 700 is close enough for government work.