More descent planning Math

Mavmb

Mavmb

Well-Known Member
http://www.boldmethod.com/learn-to-...0-to-1-rule-helps-you-plan-a-perfect-descent/

Now let's look at our second descent planning question.

We're at 10,000' MSL, and ATC gives us a crossing restriction of 5,000' for a fix that's 10 miles ahead of us. And we're flying at 120 knots ground speed.

Step 1: Our first step is to figure out how much altitude we need to lose. This is pretty easy. We're at 10,000', and we need to get to 5,000', so 10,000-5,000 = 5,000'. We need to lose 5,000 feet.

Step 2: Our next step is to figure out how long we have before we reach the fix. Since we're flying at 120 knots ground speed, we know we're going 2 MPM. With the fix 10 miles out, we'll divide 10 miles by 2 MPM and get 5 minutes (10/2 = 5). So in this scenario, we have 5 minutes to the fix.

Step 3: To finish things off we'll take the altitude we need to lose (5,000'), and divide it by the minutes to the fix (5). 5,000 feet / 5 minutes = 1,000 FPM. We'll need to descend at 1,000 FPM to make the crossing restriction.


I was taught to do this a different way? 120 is groundspeed divided by two is 600 ft a minute descent and not 1000ft in the answer?
 
I use the same 'rule' for everything:

Altitude to lose x3 = distance to start down in NM.
Groundspeed /2 = rate of descent

FL330 going to sea level? Start down 99 miles out. Doing 480 knots? Do 2400 FPM.
13000 going to 8000? 5000 feet difference is 15 miles. Doing 340 knots? Do 1700 FPM.
I usually cheat by a mile to let the airplane get established in the descent, then crosscheck it as you continue down (the checking it is the most part regardless of your formula or equipment).

It's simple and works.

Or just go VNAV D->...
 
Mavmb said:
http://www.boldmethod.com/learn-to-...0-to-1-rule-helps-you-plan-a-perfect-descent/

Now let's look at our second descent planning question.

We're at 10,000' MSL, and ATC gives us a crossing restriction of 5,000' for a fix that's 10 miles ahead of us. And we're flying at 120 knots ground speed.

Step 1: Our first step is to figure out how much altitude we need to lose. This is pretty easy. We're at 10,000', and we need to get to 5,000', so 10,000-5,000 = 5,000'. We need to lose 5,000 feet.

Step 2: Our next step is to figure out how long we have before we reach the fix. Since we're flying at 120 knots ground speed, we know we're going 2 MPM. With the fix 10 miles out, we'll divide 10 miles by 2 MPM and get 5 minutes (10/2 = 5). So in this scenario, we have 5 minutes to the fix.

Step 3: To finish things off we'll take the altitude we need to lose (5,000'), and divide it by the minutes to the fix (5). 5,000 feet / 5 minutes = 1,000 FPM. We'll need to descend at 1,000 FPM to make the crossing restriction.


I was taught to do this a different way? 120 is groundspeed divided by two is 600 ft a minute descent and not 1000ft in the answer?
Click to expand...

I think the formula you used for 600 feet/minute is one that when applied with descending 1,000 feet per 3 miles gets you approximately 3 degrees descent slope. So by that commonly used formula, 5,000 feet to lose x 3 miles per 1000' for a 3 degree descent path = 15 miles distance. Now apply your GS x 5 (or GS/2 plus add a zero, same result by a different math technique) to get 600 feet per minute, which if you do that over 15 miles, you would lose about 5,000'.

Whereas in your scenario, you have an altitude you must hit in a certain time from a certain distance which happens to be pretty close, too close for the common 3 degree descent math. So you will have to descend at a steeper angle which means a higher feet per minute. (The problem does not bring up the consideration of a descent angle either.) But as you put in your post, you know the altitude to lose, the speed and distance and from that the time to the fix. So you can just divide altitude to lose by the time you have to see exactly what rate of descent is required, and you don't worry about a nice 3 degree descent path. This problem is just about getting down on time at the right place.
 
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Or how about this:

5000' to lose, 10 miles out. That's 500 feet/mile.

You're going 2 miles/minute, x 500 feet/mile = 1000 feet/minute required descent.

But what approximate descent angle will you have?

A descent rate of 300 feet/mile is about 3 degrees. 500/300 = 1.667. So your 500 feet/mile angle is 1.667 x 3 degrees = 5.01 degrees. A good bit steeper.

500 feet/mile x 2 miles/minute = 1000 feet/minute, or about 400 feet/minute (200 more feet lost/mile x 2 miles/minute) faster descent compared to your 300 feet/mile = 600 feet/minute at a 3 degree slope.
 
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Mavmb said:
http://www.boldmethod.com/learn-to-...0-to-1-rule-helps-you-plan-a-perfect-descent/

Now let's look at our second descent planning question.

We're at 10,000' MSL, and ATC gives us a crossing restriction of 5,000' for a fix that's 10 miles ahead of us. And we're flying at 120 knots ground speed.

Step 1: Our first step is to figure out how much altitude we need to lose. This is pretty easy. We're at 10,000', and we need to get to 5,000', so 10,000-5,000 = 5,000'. We need to lose 5,000 feet.

Step 2: Our next step is to figure out how long we have before we reach the fix. Since we're flying at 120 knots ground speed, we know we're going 2 MPM. With the fix 10 miles out, we'll divide 10 miles by 2 MPM and get 5 minutes (10/2 = 5). So in this scenario, we have 5 minutes to the fix.

Step 3: To finish things off we'll take the altitude we need to lose (5,000'), and divide it by the minutes to the fix (5). 5,000 feet / 5 minutes = 1,000 FPM. We'll need to descend at 1,000 FPM to make the crossing restriction.


I was taught to do this a different way? 120 is groundspeed divided by two is 600 ft a minute descent and not 1000ft in the answer?
Click to expand...
 
Speed of 120? Isn't that called Vref? Follow the banana and decent below 10,000 is done at 250 until 91.117 says to slow down.




Sent from my Startac using Tapatalk.
 
The extra crap is too much math and mental stroking the banana.

3:1 math, 1/2 the ground speed add a zero... if you're above be aggressive and do crosscheck points to update.... back off on aggressiveness when on path.

Mental math and chill... VNAV/managed will screw you... be a pilot.
 
Beefy McGee said:
I think the formula you used for 600 feet/minute is one that when applied with descending 1,000 feet per 3 miles gets you approximately 3 degrees descent slope. So by that commonly used formula, 5,000 feet to lose x 3 miles per 1000' for a 3 degree descent path = 15 miles distance. Now apply your GS x 5 (or GS/2 plus add a zero, same result by a different math technique) to get 600 feet per minute, which if you do that over 15 miles, you would lose about 5,000'.

Whereas in your scenario, you have an altitude you must hit in a certain time from a certain distance which happens to be pretty close, too close for the common 3 degree descent math. So you will have to descend at a steeper angle which means a higher feet per minute. (The problem does not bring up the consideration of a descent angle either.) But as you put in your post, you know the altitude to lose, the speed and distance and from that the time to the fix. So you can just divide altitude to lose by the time you have to see exactly what rate of descent is required, and you don't worry about a nice 3 degree descent path. This problem is just about getting down on time at the right place.
Click to expand...
Yep.

The base (long) formula is the same. It's just algebra - you put in the variables you know to get the answer you want to calculate.
 
When ATC says "descend at pilot discretion" then you use the multiply by three rule. Or if you aren't using ATC services or if you want to do your own descent planning.

When ATC says "I need you to expedite your descent bellow ### altitude" or something, or they descend you late or early, the multiply by three rule goes out the window and you need to use alternative math.
 
36 rule. 3 x the altitude to lose, 6 x GS

3 x ALT = distance
6 x GS = FPM

Simply adjust for multiples of 3.

need to lose 10000 feet, that would normally be 30 miles but ATC in their infinite wisdom waits until you are 10 miles from the fix.
ground speed is 450 knots ( x 6 = 2700 FPM), multiply by 3 = ~ 9000 FPM.

Easy peasy, works great, last long time. Faster than plugging it into the VNAV

Want a 3 degree slope, multiply your Mach number by 3 and descend at that FPM or changer your pitch -3 degrees and maintain the same Mach number, adjust as you descend and Mach changes.
 
JeppUpdater said:
I use the same 'rule' for everything:

Altitude to lose x3 = distance to start down in NM.
Groundspeed /2 = rate of descent

FL330 going to sea level? Start down 99 miles out. Doing 480 knots? Do 2400 FPM.
13000 going to 8000? 5000 feet difference is 15 miles. Doing 340 knots? Do 1700 FPM.
I usually cheat by a mile to let the airplane get established in the descent, then crosscheck it as you continue down (the checking it is the most part regardless of your formula or equipment).

It's simple and works.

Or just go VNAV D->...
Click to expand...
That may come in handy in your near future.
 
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