Quick way to calculate ROD
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Joined: Jan 2006
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From: EU
Quick way to calculate ROD
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
descent angel or gradient and the GS...
for example:
Gradient 12%
GS 540 kt
are there any good formulas to memorize..?
/Tim
Joined: Feb 2003
Posts: 29
Likes: 0
From: Germany
What works for me
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
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
Joined: Aug 2005
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From: Belgium
Gradient versus ROD
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.
Joined: Oct 1999
Aviation Qualifications: ATPL
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From: Den Haag
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)
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)
Joined: Apr 2002
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From: Bohol, Philippines
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.
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.
Joined: Oct 1999
Aviation Qualifications: ATPL
Posts: 7,383
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From: Den Haag
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
Gradient in % x G/S kts = Required rate of descent or climb in ft/min
multiply gradient in % times G/S in kts to get RoD
Joined: Mar 2007
Posts: 378
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From: south east UK
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.
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.
Joined: Mar 2008
Posts: 38
Likes: 16
From: Victoria
ROD/GS
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.
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.
Joined: Mar 2012
Posts: 118
Likes: 36
From: Somewhere
Pilot Mental Math (Quick Reference)
🔹 ROC / ROD
Use GS × Gradient
🔹 3° Glide Path (ROD)
Use GS × 5 or (GS × 10) ÷ 2
For better accuracy, add +6% of the result
Example: GS 100 → 100 × 5 = 500
6% of 500 = 30 → ROD ≈ 530 ft/min
🔹 Mach Change
For every 0.01 Mach change, speed changes by ≈ 6 kt
( 1% of speed of sound)
🔹 Same Mach, Different Level (or same level with temperature difference)
At the same Mach, TAS varies with temperature
Warmer = faster / Colder = slower
Approx 1 kt per 1°C
Example: FL350 vs FL370 → about 4 kt difference
( FL350 -faster)
(Except inversion or tropopause conditions, no wind involved)
🔹 Different Mach + Different Level
Combine and offset the effects above
Example:
FL370 M0.84
FL350 M0.83
Mach difference = 6 kt
Temperature effect = 4 kt
→ Net difference = 2 kt
(FL370-faster)
🔹 Landing (Higher Threshold Crossing Height)
For every 1 ft above normal TCH → landing distance increases by ≈ 6 m or 20 feet
📌 Not exact values — for quick mental calculation and backup use only
Joined: Nov 1999
Posts: 647
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From: East side of OZ
In the higher levels on descent then feet per mile to lose X Mach number (which is close enough to miles per minute) should give ROD needed in feet per minute.
E.G. You need to lose 6000' in 20nms, 300 feet per nm, at Mach 0.8 you will need to decend at 2400 feet per minute.
On final approach 5 times your groundspeed.
E.G. You need to lose 6000' in 20nms, 300 feet per nm, at Mach 0.8 you will need to decend at 2400 feet per minute.
On final approach 5 times your groundspeed.
Joined: Mar 2012
Posts: 118
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From: Somewhere
Additional note on calculating ROD/ROC
The same principle can be used in all situations—regardless of altitude, including during approach.
Use feet per nautical mile (ft/NM) ÷ 60, then multiply the result by ground speed (GS) to get the required rate.(This method is more accurate and precise than using ground speed multiplied directly by the gradient.)
Example (approach):
A 3° approach is approximately 300 ft/NM
→ 300 ÷ 60 = 5
→ ROD ≈ GS × 5
(More precisely, 3° ≈ 318 ft/NM → GS × 5.3)
At higher altitude:
The same ft/NM concept applies.
Example:
Descend 6000 ft within 20 NM
→ 6000 ÷ 20 = 300 ft/NM
→ 300 ÷ 60 = 5
→ Rate = GS × 5
(500x5 =2500feet/min -high alt)
(300x5=1500feet/min -lower alt)
Another example:
Descend 12000 ft within 20 NM
→ 12000 ÷ 20 = 600 ft/NM
→ 600 ÷ 60 = 10
→ Rate = GS × 10
This method does not depend on altitude or Mach number.
If you understand it but cannot achieve the required rate, you can always use an alternative method or inform ATC.
Each pilot may have different techniques—none are wrong. It depends on what you are most comfortable using.
The same principle can be used in all situations—regardless of altitude, including during approach.
Use feet per nautical mile (ft/NM) ÷ 60, then multiply the result by ground speed (GS) to get the required rate.(This method is more accurate and precise than using ground speed multiplied directly by the gradient.)
Example (approach):
A 3° approach is approximately 300 ft/NM
→ 300 ÷ 60 = 5
→ ROD ≈ GS × 5
(More precisely, 3° ≈ 318 ft/NM → GS × 5.3)
At higher altitude:
The same ft/NM concept applies.
Example:
Descend 6000 ft within 20 NM
→ 6000 ÷ 20 = 300 ft/NM
→ 300 ÷ 60 = 5
→ Rate = GS × 5
(500x5 =2500feet/min -high alt)
(300x5=1500feet/min -lower alt)
Another example:
Descend 12000 ft within 20 NM
→ 12000 ÷ 20 = 600 ft/NM
→ 600 ÷ 60 = 10
→ Rate = GS × 10
This method does not depend on altitude or Mach number.
If you understand it but cannot achieve the required rate, you can always use an alternative method or inform ATC.
Each pilot may have different techniques—none are wrong. It depends on what you are most comfortable using.
Joined: Dec 2020
Aviation Qualifications: ATPL
Posts: 179
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From: Amsterdam
Quick way on the Bus/737 is to use the METERS (or METRES) button.
At 30,000ft, select METERS. It will show approx 9,100m. Take the first two numbers - 91.
You need 91nm plus some extra miles for speed reduction plus or minus some miles for tail/headwind.
I always round it up for sake so 100nm from FL300.
During descent, I am checking distance on progress page against my altitude using the METERS button.
At 30,000ft, select METERS. It will show approx 9,100m. Take the first two numbers - 91.
You need 91nm plus some extra miles for speed reduction plus or minus some miles for tail/headwind.
I always round it up for sake so 100nm from FL300.
During descent, I am checking distance on progress page against my altitude using the METERS button.
Joined: Feb 2015
Aviation Qualifications: ATPL
Posts: 127
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From: The black woods
Pocket descent tool
A senior captain showed me this simple method when I was a new copilot and high on descent.
It works on any aircraft with a Mach meter.
Just 2 calculations:
1. FL / Dist to go in NM = Degrees
2. Degrees x Mach Number = ROD in fpm
You can either use this in real time to assess quickly if you can make a required FL / Alt at a given point (whether you need extra drag is a judgement you then make)
Or you can use it in advance to calculate a descent point with your type standard descent angle.
It was handy on the older types where, because of cabin pressure restrictions the initial descent to FL 300 was at low rate of 750 rpm so with Mach .75, a one degree descent = 10 miles per 1000 ft,
followed by initially closer to a 4 degree descent, at Mach .7 = 2800 fpm.
Then, as the Mach number reduces in the descent, so you reduce the ROD.
Whether you need extra drag is a judgement you then make.
A handy tool and it always works, though it requires thinking outside the ‘magenta’ box.
You can use it to calculate a climb too, within performance restrictions.
Edit - see also Bullethead just above…
It works on any aircraft with a Mach meter.
Just 2 calculations:
1. FL / Dist to go in NM = Degrees
2. Degrees x Mach Number = ROD in fpm
You can either use this in real time to assess quickly if you can make a required FL / Alt at a given point (whether you need extra drag is a judgement you then make)
Or you can use it in advance to calculate a descent point with your type standard descent angle.
It was handy on the older types where, because of cabin pressure restrictions the initial descent to FL 300 was at low rate of 750 rpm so with Mach .75, a one degree descent = 10 miles per 1000 ft,
followed by initially closer to a 4 degree descent, at Mach .7 = 2800 fpm.
Then, as the Mach number reduces in the descent, so you reduce the ROD.
Whether you need extra drag is a judgement you then make.
A handy tool and it always works, though it requires thinking outside the ‘magenta’ box.
You can use it to calculate a climb too, within performance restrictions.
Edit - see also Bullethead just above…
Last edited by bill fly; 14th April 2026 at 10:08.
Joined: Mar 2012
Posts: 118
Likes: 36
From: Somewhere
Some people might wonder why different pilots use different techniques or ways of thinking. In reality, they are essentially the same—they lead to the same result. It just depends on which perspective you choose, based on what is easier and more convenient for you.
For example, comparing Bill Fly’s method and mine:
At an altitude of 6,000 ft and a distance of 20 NM, that equals 300 ft/NM.
This corresponds to a 3-degree path, because 1 degree is approximately 100 ft/NM (more precisely about 106 ft/NM).
Or you can think of it as FL60 divided by 20, which also gives 3 degrees—same as Bill Fly’s method.
For rate of descent:
• My method: GS × (300/60)
• Bill Fly’s method: TAS / 600 × 3 (mach x degree)
and also same as Mach x feet/nm
( SS is about +/- 600)
Both give very similar results.
There may be slight differences when using Mach instead of GS, especially in strong headwind or tailwind conditions(same M but not same GS).However, it is just a rule of thump and pilots already have ways to compensate for that. In the end, you can use whichever method works best for you.
PS# Alternatively, another method I use is to take the time remaining to reach a point and divide the altitude (in feet) to get the rate. For example, if point A is 10 minutes away and the altitude is 20,000 feet to go, then 20,000 ÷ 10 gives 2,000 ft/min.
For example, comparing Bill Fly’s method and mine:
At an altitude of 6,000 ft and a distance of 20 NM, that equals 300 ft/NM.
This corresponds to a 3-degree path, because 1 degree is approximately 100 ft/NM (more precisely about 106 ft/NM).
Or you can think of it as FL60 divided by 20, which also gives 3 degrees—same as Bill Fly’s method.
For rate of descent:
• My method: GS × (300/60)
• Bill Fly’s method: TAS / 600 × 3 (mach x degree)
and also same as Mach x feet/nm
( SS is about +/- 600)
Both give very similar results.
There may be slight differences when using Mach instead of GS, especially in strong headwind or tailwind conditions(same M but not same GS).However, it is just a rule of thump and pilots already have ways to compensate for that. In the end, you can use whichever method works best for you.
PS# Alternatively, another method I use is to take the time remaining to reach a point and divide the altitude (in feet) to get the rate. For example, if point A is 10 minutes away and the altitude is 20,000 feet to go, then 20,000 ÷ 10 gives 2,000 ft/min.
Last edited by Noknoipobin; 14th April 2026 at 10:40.