Hey everyone,

Something I have noticed. I have flown ILS Approaches (not sure if that matters) with the A320 probably 20-30 times at this point. Something I always notice is that no matter what airport I am going to, whether it is at sea level or thousands of feet above, ATC has me start my descent at 120 NM from the airport. I think I noticed this with the 787 as well, but don't use it as much. I was wondering if 120 miles is a realistic descent distance no matter the airport, and if others have noticed this as well.

120 miles is a realistic start point for your decent if you are at FL360.

Assuming the landing field elevation is at sea level, then simply take your cruise level, say 36000 ft, drop the zero's, so you have 36. Divide this by 3, leaving 12.

Multiply 12 by 10, giving you 120 miles.

An alternative method is to take your cruise level, drop the zero's like beforeand multiply by 3.

So 36000 becomes 36, multiplied by 3 = 108 miles.

I prefer the first method as it automatically builds in some deceleration distance to get down to 250kts below 10000ft rule.

If your landing field elevation is say 1000ft, then simply subract this from your cruise level and divide the remainder by 3.

 

Hope this helps

 

Stu

https://mediawiki.ivao.aero/index.php?title=Top_of_descent_calculation

3 minutes ago, guibru said:

https://mediawiki.ivao.aero/index.php?title=Top_of_descent_calculation

Much more in depth explanation. 

I simply divide cruise level by 3 and multiply my ground speed by 5 to find my correct decent rate.

Stu

Yeah, the FL/3 or FL*0.3 is magic number for almost all morden jet airliners.

I prefre *0.3 because is more easy for mental calculation

If you need more accurate calculation, use the FMC then (assuming it's properly simulated then....)

22 minutes ago, Bunchy said:

Much more in depth explanation. 

I simply divide cruise level by 3 and multiply my ground speed by 5 to find my correct decent rate.

Stu

Why use 3 for distance and 5 for rate? Both those numbers in the calculation represent the angle of descent... correct?  Why not use the same angle for both?

The distance calculation is using the descent angle in degrees.

The speed calculation is using the TAN of the descent angle.

 

1 hour ago, Virtual-Chris said:

Why use 3 for distance and 5 for rate? Both those numbers in the calculation represent the angle of descent... correct?  Why not use the same angle for both?

Because groundspeed dictates how many feet per minute you need to achieve to maintain a 3 degree decent rate.

Hi.

I'm not good at mental math so I use this :

https://x-planetools.com/descent.html

but it's true that it's always good to know the math formula.

See you 

 

Edited November 5, 20205 yr by OlivierCanada

8 hours ago, Bunchy said:

Because groundspeed dictates how many feet per minute you need to achieve to maintain a 3 degree decent rate.

I know, but the constant used (3 or 5) is the desired glide slope for the descent. Why do you use 3-degrees for your descent distance calculation but 5-degrees for your decent rate calculation? Or do I misunderstand what the 3 and 5 represent?

Edited November 5, 20205 yr by Virtual-Chris

11 hours ago, Bunchy said:

I simply divide cruise level by 3 and multiply my ground speed by 5 to find my correct decent rate.

This is the easiest and works great for VFR when you have to figure when to start your descent.

Although, I divide by 300 (24,000 feet to lose / 300 = 80 miles out for top of decent - VFR example 2500 feet to lose / 300 = 8 out or so miles top of descent better to round up to 9, makes sure you’re near pattern altitude a mile or two out)

Interesting, I also found a website that had the 3-degree method where you multiply by 3. I actually wrote a quick program in Java to test it out quickly without having to mess with a calculator every time and I will be darned its always about 110-120. Cool! I have the cruising altitude-airfield elevation x 3. Seems fairly consistent.

https://www.thinkaviation.net/top-of-descent-calculation/

Edited November 6, 20205 yr by M3Stang

15 hours ago, Virtual-Chris said:

I know, but the constant used (3 or 5) is the desired glide slope for the descent. Why do you use 3-degrees for your descent distance calculation but 5-degrees for your decent rate calculation? Or do I misunderstand what the 3 and 5 represent?

To descend at 3 degrees you have to lose 1000 ft per 3 miles. The faster you go, the more height you need to lose in a given time.

If you are flying at 450kts ground speed, as an example, you are covering the ground at 7.5 nm per minute. So from this you can see you would have to have a decent rate of around 2500ft a minute, to be on that 3 degree slope.  (It's rule of thumb to keep the mental calcs simple.) We don't use 5 degrees, we use 5 as a multiplier of ground speed.

If you multiply 450 by 5, that gives you a decent rate of 2250ft per minute, very much in the ball park of 2500ft per min.

You can also half your ground speed and multiply by 10 to arrive at the same result.

450 / 2 = 225 * 10 = 2250.

The important thing to remember when working out decent rate is to use ground speed, not true airspeed or indicated airspeed.

 

Makes sense?

 

Stu

Edited November 6, 20205 yr by Bunchy

Just load the STAR and/or approach into the FMS, the altitudes will be pulled from the navdata and then the system will produce a TOD for you. No old school fancy math needed, let the airplane do what it is designed to do.

19 hours ago, Virtual-Chris said:

I know, but the constant used (3 or 5) is the desired glide slope for the descent. Why do you use 3-degrees for your descent distance calculation but 5-degrees for your decent rate calculation? Or do I misunderstand what the 3 and 5 represent?

The 5 in the rate calculation is NOT 5 degrees. it is Tan 3 degrees which is 0.05 or 5%