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

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

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

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.

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

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! :)

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.

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)

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.

SFI145 Well said!

Ah, yes that's entirely different to what I said::ugh:

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.

ROD/GS
 

Or for Descent gradient =ROD/GS or for Climb gradient= ROC/GS to find the FPM you need to make the gradient required.....

Ten times your groundspeed divided by two

EG 160 times ten equals 1600 divided by two for an 800 fpm vs on approach

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

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.

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.

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.

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…

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.

I think you mean 5.03. No?

Thank you ,I mean it’s actually 5.3, not 5.03.

tan 3° = height / distance
0.0524 = X / 6076 (1 NM)
→ X = 6076 × 0.0524 ≈ 318.4 ft

So, over 1 NM the height is about 318 ft.

Make 1 NM = 1 minute then→ GS = 60 kt
if flying 60 kt → ROD ≈ 318 ft/min
If flying at Y kt: ROD = (318 / 60) × Y ≈ Y × 5.3

When we truly understand the actual values, it helps improve our situational awareness and caution.
In real operations, it may be difficult to fly exactly at an ROD of 530 (GS × 5.3)(in V/S mode). However, if we are flying at around 500 (GS × 5), we can still anticipate how the aircraft profile will develop. This helps avoid confusion or surprise about why, even with a “correct” ROD (GSx5), the aircraft may appear high.

Try looking at an example AIP chart for EGLL.



https://cimg5.ibsrv.net/gimg/pprune....b34807b91f.png

Wondering how I ever got through a 43 year pilot career without these formulae ringing in my head…:ugh:

Right you are. I am embarrassed at my slip and at my gall in offering a correction.

I was just thinking the same thing. People seem to be making an easy job, very hard!

I understand your sarcasm. I’m simply explaining this for those who wonder if what I wrote is right or wrong, providing a clear source and reference so it’s not just guesswork. Besides, this isn't difficult at all—it’s just middle school or even upper elementary math. The entrance exams to become a pilot were far harder than this. I’ve been flying for over 30 years and have seen many pilots who still can't operate an FMS and have to rely on their co-pilots, or those who can barely read a chart—yet they’ve all managed to survive or reach retirement.

I’m sorry for making you feel that way. I was only trying to explain and share a bit more about how I work,so that it’s credible and accurate.Sometimes I may speak too much.

Not at all. I appreciate your explanation. My mistakes are entirely my own fault.

Well, of course you can get in from a sub optimal approach at the cost of fuel but its nicer to hit it on the button. I was grateful for that tip from my chief and have used it on countless occasions.

Apart from approaches, in the crowded airspace of today you often get asked if you can make a given height (climbing or descending) by a given point or distance.
Then its nice to have a handy tool to give a quick answer.

Handy being the operative word though. As a slower learner I have always appreciated the simple solution and I agree that some of the examples look anything but at first sight.