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# Rationale of the altitude*3 theory

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# Rationale of the altitude*3 theory

8th Jul 2012, 15:06

Join Date: Apr 2008
Location: Test
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Rationale of the altitude*3 theory

Hi there,

Can anyone point me in the correct direction of the explanation of this theory. Or if time is in your side, care to give an explanation?

I know it's something to do with the glidepath of 3deg. Assuming we are at 4500ft, we must be at least 13.5NM away from the airport. What do we mean by "13.5NM away from the airport"? Does it mean that we must be at least 13.5NM to intercept the LOC or 13.5NM to descend?

What if im at 13.5NM away, but the purple dot is on the top? Shouldn't it be on the correct glidepath?

Anyone care to explain?

Thanks.
8th Jul 2012, 15:30

Join Date: Sep 2010
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3 times track miles = FL.
Works well up to Fl 300, above that you achieve a greater ROD per NM.
Below Fl 100 with speed control then slightly flatter.
But if you steam down at MMO/VNE and have a friendly controller then leave cruise level much later. As in the good old days and sod the passengers eardrums.

Example if you are at FL 270 start the descent at 90 travel miles. You will also need to factor in wind component.
Hugh Dibley invented a circular slide rule descent calculator in the 70s.
8th Jul 2012, 15:43

Join Date: Nov 2007
Location: Texas
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While gliders have made significant advances over the last 50 years, airliners seem stuck with an 18 to 1 glide ratio. But this makes the math easy.

If you are 4500 AGL you should have 13.5 track miles to the airport. Plus some to slow and configure and +/- some for the wind. It's all well and good to figure all this out but ATC will most times vector you around and screw up whatever plans you or the box had.
8th Jul 2012, 15:59

Join Date: Dec 2011
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Intercepting the LOC is irrelevant to that.
I am not sure how many miles the reception of the LOC goes out to, but I remember 17 miles from flight school (could be wrong).

The x3 is basically because on a 3 degree glide path you are losing about 300 feet per NM.
So if at 4000 feet you were at 12 miles, u are more or less on the glide path.
If at 4000 feet u were at 9 miles, u know ure above profile and have to capture the GS from above.

Some guys multiply NM by 3.
Some multiply FL by 3.

Give a little different numbers. Would like to know myself which one is better used.

Last edited by WhySoTough; 8th Jul 2012 at 16:00.
8th Jul 2012, 16:12

Join Date: Aug 2009
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(Altitude x 3) OR (Distance x 3) comes from the fact that the ideal descent path of an aircraft is 3 degrees. Further, most ILS approaches, if not all, have a glide path angle of 2.5 to 3.25 degrees.

If you use 3 degrees as an average, what you will find using trigonometry is that for a given distance (NM), multiplying that distance by 3 will give you the height in FT. As an example, let's use a 3 degree profile and a ground distance of 10NM.

Tan(3) = Height/10NM

Tan(3) x 10NM = 0.524NM = 3183ft

As you can see, we need to make approximations and have to round numbers up or down - but even with the approximation and simple maths in your head, you will see that in real life it works quite well.

Using trigonometry you can see that the (Altitude to lose x 3) to give you distance required works quite well also.

Now this distance/altitude relationship is directly to touchdown. For instance, if you are 10NM away, you should be at 3000ft and descending at the appropriate descent rate to achieve the 3 degrees from your present position to touchdown. But as some of the others have already mentioned, you need to take into account your airspeed as well as the headwind/tailwind component. For instance, it is no good being at 3000ft at 300KTS and with a 20KT tailwind. Technically, you are still on the "correct" glide path as calculated but given the situation, it is unlike that you will achieve a desirable outcome.

This x3 method doesn't just work for distance or altitude to touchdown. It also works for meeting altitude restrictions on your way down. For example, (height to lose x 3) will give you the distance from the waypoint in which you need to begin your descent. Equally, checking if you are high or low to meet a altitude restriction can be done by using (Dist to waypoint x 3).

You will find that these are general rules of thumb, and are not meant to be exact. But they work surprisingly well and using these methods will bring you pretty much on profile when you join the glide slope.

8th Jul 2012, 16:22

Join Date: Jan 2006
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3° path is 300ft/nm, 10nm 3000ft, 15nm 4500ft height

multiplying fl by 3 is useful fl 400 give you TOD 120 nm
but 120 nm by 3° give you 36000

the difference is speed deceleration

speed fl 400 to fl 100 let s say 300 kt
fl 100 250 kt to plateform 200 kt

so speed reduction is 100 kt so 10 nm level flight

120nm by 3° give you 36000 + 10 nm level = 3000ft= 39000ft so quite close than FL 400* 3= 120 nm

another interesting thing is ISA temperature deviation

G\S is a ground slope, at 10 nm you are 3000ft height you are good in ISA 0
isa -25 your altimeter is overeading your true altitude at 3000 indicated your are in fact 2700 ft you are low
in isa 25 your altimeter is underreading your true altitude so 3000ft indicated give you 3300 ft true altitude your are high

in the same manner a VNAV approch 3° in isa 25 become a airslope 3,3° so no surprise to discover that you end up with 3 white 1 red on the PAPIs because Papis are ground slope too

does it make sense

Last edited by tony montana; 8th Jul 2012 at 16:23.
8th Jul 2012, 16:50

Join Date: Feb 2008
Location: Bellingham WA
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The arithmetic

Since this is a rule of thumb, somebody chose a number close to a desired result. The example above, 90 NM and FL 270, tells us the desired result, which is the glide slope (G). It's a right triangle, so tan(G) = height / distance.

Using 6,076 ft per NM, the height (H) is FL 270 * 100 = 27,000 / 6076 = 4.44 NM, so tan(G) = 4.44/90 = 0.049, and the glide slope is G = arctan(0.049) = 2.83 degrees. Close enough to three degrees for a rule of thumb.

with D always in NM, if we want a general answer in NM, it's easy: solving tan(G) = H/D, H = tan(G)*D = 0.049*D NM.

To get to feet, H = 0.049 * 6,076 * D = 300 * D.

To get to FL divide by 100, and H = 3*D.
8th Jul 2012, 17:06
Per Ardua ad Astraeus

Join Date: Mar 2000
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All of that dreadfully complicated and unnecessary. Tangents??

1 in 60 is your friend - heard of that? By the way, stick with Altitude/3 - it is far better!
Just stick with that. You can do the maths if you wish, remembering 1nm is near enough 6000ft. but much simpler to learn the 3 times table.

Thank the Lord for nautical miles What those strange Europeans do I do not know.
8th Jul 2012, 19:14

Join Date: Feb 2008
Location: Bellingham WA
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My apologies to the OP - I answered the wrong question. The 3*thousands-of-feet rule gives about a 3.1% glide slope, which would help to intercept from above. The 3*distance (2.8 degrees) rule would help to intercept from below.

BOAC:
My abject apologies. My poor little pea brain is incapable of realizing the H = 3*D can be written as D = H/3. If you ARE using altitude, though, I sincerely hope you are not above mountains.

- Your humble and obedient servant
8th Jul 2012, 20:36
Per Ardua ad Astraeus

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That's ok, Michael, your brain is excused. My post was for the OP.
8th Jul 2012, 21:21

Join Date: Jan 2006
Location: toulouse
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to calculate TOD use FL*3, rule works until you start decelerating to 250kt
to calculate correct altitude final stage distance*300= target altitude at up speed
9th Jul 2012, 06:58

Join Date: Mar 2005
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It depends on the particular airplane, weight, ias, wind...

I use FLx3 plus 10 (only plus 6 below 10000) for typical weights and speeds, and when I am in VNAV, i find that rule to be very accurate, specially at lower levels where weight or speed or wind make less difference.

When on the magenta donut, with 8000 i always have about 30 to go, with 6000 always 24... And so on. Just as my rule says
9th Jul 2012, 09:01

Join Date: Oct 2000
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The currently recommended figure of three times the number of thousands of feet was arrived at through extensive TE research and is now a JAR-compliant standard for most types. In addition, it fulfils the requirements of ETC and CE. It will, provided you apply the corrections for external conditions laid down by YOE, remain within the acceptable limits of the ITBP tolerances.

I feel that a few of the acronyms may warrant an explanation:
TE – Trial and Error
JAR – Just About Right
ETC – Easy To Calculate
CE – Close Enough
YOE – Years Of Experience
ITBP – In the BallPark

Last edited by ft; 9th Jul 2012 at 09:03.
9th Jul 2012, 09:22
Per Ardua ad Astraeus

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Sorry, ft - I must point out that you missed 'TLAR' which is a vital component of the calculation,.
9th Jul 2012, 09:25

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D fell of my chair there... Good one
9th Jul 2012, 13:17

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Just a suggestion, try use the meters scale and the trackmiles will be a direct readout! Unless u really fancy exact calculations.. Add a few to allow for slow down, another few if tailwind and so forth..
9th Jul 2012, 20:01

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Thanks guys.

Say im at 4500ft with about 20NM to go. What is your way of determining the descent rate for the aircraft?
9th Jul 2012, 20:16

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There's no way of answering that without knowing your groundspeed.

OTOH being pragmatic and a simple soul, and a believer in the KISS principle if pushed I'd just fly level until intercepting the 3 degree slope ( at about 15 miles) then fly the 3 degree slope, with an ROD of approx 5 X ground speed)

Last edited by wiggy; 9th Jul 2012 at 20:21.
9th Jul 2012, 21:07

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But if you do have access to your groundspeed, which is pretty typical in the GPS era, it's a simple calc:

Rate of descent to lose 4500ft in 20nm?
1) Check groundspeed. Let's say 360kts for this example.
2) 360kts = 6nm / min, therefore 20nm = approx 3 mins
3) 4500ft / 3 mins = 1500 fpm required descent rate

It's better to round down at step 2, whether climbing or descending, to ensure that you reach your required altitude before using up the specified distance. This method can be used during a climb to quickly answer questions from ATC such as "can you be at FLXXX by point YYYYY", although you need to make a sensible assessment of how much your rate of climb will diminish as altitude increases.

Last edited by Easy Street; 9th Jul 2012 at 21:09.
10th Jul 2012, 07:05

Join Date: Mar 2005
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Just check if distance to go is good for your height.

The rate varies a lot when you make idle descents. But, for 3 deg slope, If you want to calculate if you are making 3 deg or not, try this, using GS:

300 kt aprox 1700 fpm
250 kt aprox 1500 fpm
200 kt aprox 1000 fpm

I use aprox GS in nm per min times 3, add two zeros and plus a 10% of that (it's like multiplying by 330)

Example.
At 300 kt you fly at 5 nm per min, so in a min, at 3deg (330 fpnm) you should descend 5x330=1650 ft.

I use it all throughout my descent so I know if I am progressing as planned or not. I have noticed that the average gradient keeps fairly constant, so I can extrapolate and predict the trend, so I know early if I will need higher speed or speedbrakes or maybe reduce rate for a while if Im getting low.

I know I make too much maths, but I like it. i also monitor still air gradient, with TAS. which helps me check the effect of weight and speed in the gradient