Rule of thumb help
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Rule of thumb help
hello everyone, Just wondering if anyone knows a quick rule of thumb to solve this kind of problems:
Descending at X fpm with X ground speed what distance over the ground would you cover descending from X to Y altitude?
And distance Vs altitude on 3° glide with a visual vertical guidance during the approach.
many thanks
Descending at X fpm with X ground speed what distance over the ground would you cover descending from X to Y altitude?
And distance Vs altitude on 3° glide with a visual vertical guidance during the approach.
many thanks
Join Date: Apr 2010
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1. If you divide difference in altitude by rate of descent, you get time in minutes. You then multiply this by ground speed, dividing by 60 (since ground speed is in knots = NM/h).
Track distance = (Altitude difference / Rate of Descent) x (Ground Speed / 60)
For example, you have to descent 10000ft with ROD 2000ft/min and you have ground speed of 300 knots. Dividing 10000 by 2000 will get you 5 minutes. If you divide 300 knots by 60, you get 5 NM/min. If you multiply 5x5, you get 25 NM - very simple indeed.
2. The rule of thumb for 3° glide slope is very simple, you multiply difference in altitude (let's say from present altitude to runway elevation) with 3 and you get distance.
Distance [NM] = Altitude Difference [ft] x 3
So if you are 4000ft above the runway, you should be 12 NM from the threshold. If you're closer, you're coming high and if you're further away, you're shallowing
Track distance = (Altitude difference / Rate of Descent) x (Ground Speed / 60)
For example, you have to descent 10000ft with ROD 2000ft/min and you have ground speed of 300 knots. Dividing 10000 by 2000 will get you 5 minutes. If you divide 300 knots by 60, you get 5 NM/min. If you multiply 5x5, you get 25 NM - very simple indeed.
2. The rule of thumb for 3° glide slope is very simple, you multiply difference in altitude (let's say from present altitude to runway elevation) with 3 and you get distance.
Distance [NM] = Altitude Difference [ft] x 3
So if you are 4000ft above the runway, you should be 12 NM from the threshold. If you're closer, you're coming high and if you're further away, you're shallowing
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I don't know if this will help as it's not a particular rule of thumb but...
Nothing wrong with drawing out the triangles and converting the units. I would guess that for your first example, where there are 5 variables, of which there are presumably four given and one unknown, it's unlikely that there is a rule of thumb. They normally only work when certain things remain constant, not when everything is variable, so for these I reckon you'll just have to work them out.
Your second example you are much more likely to get a rule of thumb for, as the 3 degrees component stays constant.
At 3 degrees, for every 2 miles there is a 545ft drop.
At 5 degrees, for every 2 miles there is a 10416ft drop.
At 10 degrees, for every 2 miles there is a 1836ft drop.
These were worked out by trigonometry, using
tan a = opposite/adjacent
with the adjacent being the distance (in feet, 1mile is 5208ft), and the opposite being the height difference.
Conclusion - trigonometry is probably the area to be looking at.
Any good for you?
Edit - FS wrote at the same time that I was, and is making a lot more sense than I am!
Nothing wrong with drawing out the triangles and converting the units. I would guess that for your first example, where there are 5 variables, of which there are presumably four given and one unknown, it's unlikely that there is a rule of thumb. They normally only work when certain things remain constant, not when everything is variable, so for these I reckon you'll just have to work them out.
Your second example you are much more likely to get a rule of thumb for, as the 3 degrees component stays constant.
At 3 degrees, for every 2 miles there is a 545ft drop.
At 5 degrees, for every 2 miles there is a 10416ft drop.
At 10 degrees, for every 2 miles there is a 1836ft drop.
These were worked out by trigonometry, using
tan a = opposite/adjacent
with the adjacent being the distance (in feet, 1mile is 5208ft), and the opposite being the height difference.
Conclusion - trigonometry is probably the area to be looking at.
Any good for you?
Edit - FS wrote at the same time that I was, and is making a lot more sense than I am!
Last edited by 4015; 21st Jun 2011 at 11:26.
