ROD % in Fpm

[wilcoluca]

How do you convert the ROD in percentage into fpm?

[DBate]

Multiply the % gradient by 60 to get the number of feet per NM.

Example: 6% Gradient
6 x 60 = 360 ft per NM

[Old Smokey]

I don't think so DBate, not unless your Ground Speed is 60 Knots.

wilcoluca, I think you'll get much better results if you use -

ROD (fpm) = % Gradient X Ground Speed (Knots)

e.g. for a 120 Kt Groundspeed and a 6% gradient, ROD = 6 X 120 = 720 fpm. (For the purists it's actually 729 fpm, but who can read 9 fpm).

Regards,

Old Smokey

[JABBARA]

Dbate,

Your equation is correct to onvert the gradient value from ft/NM to Percantage or vice versa.

Example 300 ft/NM gradient is eqaual to 5% (=300/60);

Or

8 % gardient equals to 480 ft/NM (= 8 x 60)

These equations can easily be derived because 1 NM equals approximately 6000 feet.

And one more small thing: For small angles, 1.7 % gradient equals approximately 1 degrees (Because Tangent 1 degree equals to 0.017). So If you have FPV ( bird) on the EADI, you can read your instantenous climb gradient.

Regards

[DBate]

Old Smokey ,

your formula is of course right when you are looking for a way to convert a gradient into the required fpm.

And of course yours is the correct answer to wilcolucas question.

But , if you take a close look, the formula I gave is one to get the feet per NM, not the fpm.

So picking up your example with 120kt GS:

120 kt GS ---> 2 NM/min
Gradient of 6% ---> 360 ft/NM

Multiplying 360 ft/min by 2 gives the fpm.

So, I admit, you gave the correct answer to the original question. The formula I gave is nevertheless correct. :ok:

Cheers,
DBate

Edited to say that JABBARA was by a minute faster with his reply

[Rainboe]

DBate, it still doesn't work. A jet doing about 145kts on final approach on about a 3 degree slope needs a ROD of about 800 fpm. According to your formula, 3% x 60= 180 feet per mile. At about 150 kts, that's 2.5 x 180=450 fpm which is way too low.

The only formula I ever bothered with on a 3 degree slope is 5 x GS as a rough guide. Being lazy, I didn't do a calculation. Just set a baseline of 800 fpm and adjusted it up or down for slope, unusual GS etc. Why fill your mind with all that jazz when you have other things to concentrate on?

[-IBLB-]

how about in degrees:

ROD(ft/min) = 1.75 x glidepathangle(degrees) x GS(kts)



-IBLB-

[alf5071h]

Do Jeppesen still publish a table of gradient to rate and vice versa in the approach chart documentation? If so, why not use that, much simpler in a stressed or time constrained situations, should also reduce error (and debate, as interesting as it is).

[-IBLB-]

alf5071h Yes, JEPPESEN publishes it somewhere in the beginning of the enroute section i believe. NACO publishes it aswell, on the inside of the back-cover page.

For operational use i would definitely suggest that aswell.

wilcoluca I was wondering why you want it from percentage -> descentrate, not angle -> descentrate? Usually on charts the angle is published, not the percentage.

-IBLB-

[F/O]

@ Rainboe
percentage and degree are not the same....!
As for the last part of your post I agree: Use whatever is easy for yourself and this can be different for everybody; a ballpark of 800 fpm, 5x GS, use of a table or whatever other formula or method

[JABBARA]

Rainboe,

As -IBLB- says

3 degrees glide slope equals to 3 x 1.7 = 5 % gradient.

With 150 kts (let's say Ground Speed) to remain on glide path the required vertical speed equals 5 x 150 = 750 fpm.

Regards:ok:

[Rainboe]

Woops- mixed my degrees and gradients up! Whoever works in gradients? The only reference I have ever seen to them being used practically is climbout angles in high terrain areas.

[enicalyth]

I mean climb or descent in percentage, say h%, is merely h feet vertically in 100 feet horizontally. It is schoolboys and schoolgirls the tangent of the angle "a" =h/100.

But your vertical speed and horizontal speed must be in the same proportion. Little v for vertical, Big V for horizontal.

Therefore h/100 = v/V = tan "a" where both v and V are measured in the same units.

Therefore v = V tan "a".

Committing the gross sin of assuming a nautical mile is 6080 feet
v = (Vh/100) * (6080/60) = Vh*(6080/6000) = 1.013333Vh

which is nearly Vh in my book.

ROD fpm = GS knots * %age gradient.

At 240kts and 3% is that 720fpm? At 120kts and still 3% is that 360fpm?

I theng you

I'll get my coat.....

[Old Smokey]

I've never seen so much talking in circles over a fairly basic matter. May I endeavour to provide 3 practical conversions for day to day use?

(1) Calculation of Rate of Descent for a given Gradient in %

ROD (fpm) = % Gradient X Groundspeed

(2) Calculation of Rate of Descent for 3° Glide Slope

ROD (fpm) = 5.3 X Groundspeed (At Jet speeds, 5 X G/S + 50 fpm)

(3) Conversion of % Gradient to Flight Path Angle (FPA) degrees

FPA = % Gradient X 0.6

Simple isn't it?

enicalyth, surely a man of your calibre would use 1 nm = 6076.1155' (just being my pedantic self).

Regards,

Old Smokey

[Rainboe]

I tend not to be greatly interested in aviation related calculations I can't do in my head whilst instrument flying with one hand and 90% of available brain space pre-occupied with the flying! It's all very well doing these bizarre calculations, but the flying and keeping a lookout come first, and frankly, converting gradients to fpm at current groundspeed is one of those esoteric maths problems there is no room for!

Still shouldn't have mixed up my units there though. Must be getting old.

[enicalyth]

Old Smokey! G'day!!

I love pedants, being one of those damned pedantic captains!

But I have never understood chart making and makers.

If the object of their exercise is to determine the circumference of an ellipse then
C = pi (a + b) [ 1 + 3 h / (10 + (4 - 3 h)^1/2 ) ]
where h = (a - b)^2 / (a + b)^2 .

Assuming that a = 6378137 metres and b = 6356752.314 metres
Circumference = 40007862.92 metres. So if a sphere had the same circumference the nautical mile might be 1852.216 metres or 6076.824 feet… but it isn’t.

If on the other hand their object of the exercise is to determine the surface area of a spheroid
Then

S = 2 pi a^2 + (pi / e) b^2 ln (1+ e)/(1 - e) where e = eccentricity = (1 - b^2 / a^2)^1/2

S = 5.1066E+14 square metres and a sphere of the same surface area would have radius 6371007.181 metres which gives us a possible nautical mile of 1853.251 metres or 6080.22 feet which it used to be, ish, and probably still is to Jeppesen and the US DoD who happily tell us in inches to the nautical mile.

Now wise men will tell you that the spheroid “radius” at latitude +/- 45 degrees is as near as dammit 1852 metres or 6076.1155 feet. So be it. We obviously fly further now for the same amount of fuel which will please the accountants.

But if you want the nearest possible conformal projection of a spheroid onto a sphere, 6080 feet it is for a nautical mile and a radius of 20901500 feet. But as you rightly point out, I am by International Definition, out of step.

Oh the heresy of it all, Old Smokey!!! I am wrong and everbody else is right!!!!

[They've taken the laces out of my shoes and I'm not allowed anything sharp you know. But the room is nicely carpeted... on the floor, the walls, the ceiling... ha ha!!!!!!]

You'll be pleased to know retirement beckons.

Best Rgds in all good humour

enicalyth

[Piltdown Man]

Just keep to 3 degrees or so and use your groundspeed, divided by two and then add a zero). For example, 180 Kts and a 3 degree slope gives you the following: 180/2 = 90 plus a "zero" equals 900 feet per minute. Extra hard sums are those that clever people do on the ground to impress others. Us simple pilots do easy sums, especially when flying.

[rigpiggy]

exactly, for 3 degree slope figure 3 timesyour altitude for descent. ie: @4000' TDZE is 800asl=3200'*3= 9.6 nm from threshold set 1/2 your G/S +a zero ie: 160kts 800fpm, 130 kts=650fpm. approximates to be sure, but then again I can only hold +/- 5 kts, degrees, or +/-100' sometimes close enough is good enough, other times use the damn box

[BOAC]

Rainboe - converting gradients to fpm at current groundspeed is one of those esoteric maths problems there is no room for - lucky you did not use Jepps, then! FYI, approach gradient is published on Jepps - but not, IIRC on Aerad or whatever they are called today. Therefore it IS of great practical interest to aviators.I've never seen so much talking in circles over a fairly basic matter -OS - its all matter of degree:D


enicalyth - have a happy retirement:D :D

[Old Smokey]

BOAC - It took a while, but I see your angle.

[Rainboe]

I'm finding this very 10% uphill!

[Old Smokey]

Yes Rainboe, those 6° slopes can be hard going.:cool:

[BOAC]

You are all being quite obtuse and I have an acute pain somewhere. I think my reflexes are not all they should be.