CLARKGRSWOLD
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- Joined
- Sep 28, 2007
- Posts
- 278
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- If you are 200 miles out @ 310, ATC wants 110 @ 40 DME, when do you descent ( I WAS THINKING 80 MILES BEFORE THE 40 DME, ANY INSIGHT ? )
- At 2000 feet on localizer, 5 miles back, where do you expect glideslope to be (This one is confusing 5 miles back from where ? Typical 3:1 here however )
Just use the banana bar. Sheesh.
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- If you are 200 miles out @ 310, ATC wants 110 @ 40 DME, when do you descent ( I WAS THINKING 80 MILES BEFORE THE 40 DME, ANY INSIGHT ? )
- At 2000 feet on localizer, 5 miles back, where do you expect glideslope to be (This one is confusing 5 miles back from where ? Typical 3:1 here however )
Glide slope will be above you as you must intercept below the GS due to a possible side lobe of the 90/150 Hz amplitude phases.
[/QUOT
- If you are 200 miles out @ 310, ATC wants 110 @ 40 DME, when do you descent ( I WAS THINKING 80 MILES BEFORE THE 40 DME, ANY INSIGHT ? )
- At 2000 feet on localizer, 5 miles back, where do you expect glideslope to be (This one is confusing 5 miles back from where ? Typical 3:1 here however )
Glide slope will be above you as you must intercept below the GS due to a possible side lobe of the 90/150 Hz amplitude phases.
What? What college are you pretending you went to?
Thousands of feet to descend: 20
Times 3
=60 miles out from fix
Ground speed times 5
GS:500kts
=60 miles out at 2,500fpm
Somewhere, a 727 captain is rolling in his grave at the loss of the mental 3:1 descent planning calculation...
Your math is easy and works well. However, if you want to be even more precise (and be able to explain why the math works), divide the altitude you want to lose by 300. (In the example, you get 20,000/300=67 miles out). This is because if you are descending on a 3 degree glidepath, you are descending 300 feet down for every 1 NM you go forward. Hence, the altitude to lose divided by 300 feet gives you the miles it would take to lose that altitude.
VNAV BabyBut yah, take the number of feet to lose times it by 3
At FL Level 350 / Cross X Fit At 15,000 feet = 20,000 feet to descent (20 x 3 = 60), Also, you might want to add 10 miles for the initial pitch down and level off delay. Than, take the GS and divide it by 2. So 500 KIAS / 2 = 250 (add a 0) = 2500 FPM.
Wow. What airplane are you flying that does 500 knots indicated?![]()
Point nose down, go down.
- If you are 200 miles out @ 310, ATC wants 110 @ 40 DME, when do you descent ( I WAS THINKING 80 MILES BEFORE THE 40 DME, ANY INSIGHT ? )
- At 2000 feet on localizer, 5 miles back, where do you expect glideslope to be (This one is confusing 5 miles back from where ? Typical 3:1 here however )
It's not as hard as it first looks. Here's how to do it easily (and impress your captain/fo at the same timeDividing numbers that large in my head makes me feel ill.
2. Your high
It's not as hard as it first looks. Here's how to do it easily (and impress your captain/fo at the same time):
Everyone knows that certain numbers are easily divisible by 3 (3,6,9,12,15,18,21,24,27,30,etc.). If you have to lose 18,000', it's 60 miles; if you have to lose 21,000', it's 70 miles. But what if your altitude is in-between those numbers? Notice that there are always only two choices (in this case 19,000' and 20,000'). The lower of the two choices will be 63.333 (call it 63); the higher of the two will be 66.667 (call it 67). The lower number will always end in 3, the higher will always end in 7.
Try another example: 13,000'--it's closest to 12,000', so the distance will be 43 miles. 14,000'--it's closest to 15,000', so the distance will be 47 miles. This always works, and with a little practice you can look at a number and tell instantly what the correct distance is.
This method is especially useful for the precision necessary for approach descent planning, particularly the op's original question about being at 2000' and 5 miles out--where is the G/S? At 2000', that's closer to 2100' (which would be 7 miles) than 1800' (which would be 6 miles), so you need to start down 6.7 miles out. If you indeed are at 2000' at 5 miles, you're WAY high! (You should be at 1500' at 5 miles). If you're still confused, just ask the guy sitting next to you.
Last Transmission was broken and stupid.
That's okay. Some pilots like to improve their mental abilities with age. Others prefer to be RJLosers their entire life. Take your pick.
It's not as hard as it first looks. Here's how to do it easily (and impress your captain/fo at the same time):
Everyone knows that certain numbers are easily divisible by 3 (3,6,9,12,15,18,21,24,27,30,etc.). If you have to lose 18,000', it's 60 miles; if you have to lose 21,000', it's 70 miles. But what if your altitude is in-between those numbers? Notice that there are always only two choices (in this case 19,000' and 20,000'). The lower of the two choices will be 63.333 (call it 63); the higher of the two will be 66.667 (call it 67). The lower number will always end in 3, the higher will always end in 7.
Try another example: 13,000'--it's closest to 12,000', so the distance will be 43 miles. 14,000'--it's closest to 15,000', so the distance will be 47 miles. This always works, and with a little practice you can look at a number and tell instantly what the correct distance is.
Okay, rather than arguing, "you're simply wrong" back and forth, let's do the math and put this question to bed. I had always been told that you descend 300' per nautical mile in a 3 degree descent. If this is true, then my way would be the most precise, and your way would be "close enough." But let's see.3 X the altitude = distance.
18k=54 miles
21k= 63 miles
1 mile for 10 knots airspeed to lose.
You are making stuff way too hard and simply wrong.
You're under by 3.25 miles; I'm over by 2.75 miles. Therefore, my method is a tiny bit more precise (and a little more conservative for those who don't want to bust an altitude), but it's not "simply wrong." Bottom line, use whichever method you want; there's nothing wrong with discussing why each method works, though.