rates of descent
Join Date: Mar 2000
Location: England
Posts: 97
Likes: 0
Received 0 Likes
on
0 Posts
Everyone is getting too deep here. let's RTFQ and give a simple answer.... IT'S THE LIGHTER ONE.
In order to maintain the same speed with idle thrust the lighter aircraft will have to increase pitch down resulting in a higher rate of descent. Those of you who have operated empty sectors in jets compared to a full load will have already experienced this many times, and some very high descent rates can be achieved in a light jet aircraft of same type at same speed as a heavy one provided thrust remains at idle.
KISS
"two jet a/c's of the same type, descending at the same speed, throttles are at idle, one significantly lighter than the other, which will have a higher rate of descent?"
KISS
Join Date: Oct 2000
Location: N. Europe
Posts: 436
Likes: 0
Received 0 Likes
on
0 Posts
Take a DC-9-41. In a speedbook I have handy I find a maximum listed weight of 52t and a minimum weight of 30 t.
Doing a bit of algebra, I find that the rate of descent is proportional to the drag (D) through the mass (m) of the aircraft. The drag will of course in turn depend on the mass, D(m).
Thus, the difference in descent rate, heavy aircraft descending faster being the positive, can be described by
V/S difference = D(52)/52 - D(30)/30 = ( 30*D(52)-52*D(30) ) / (52*30)
In other words, if
30*D(52)-52*D(30) > 0, the heavy aircraft will descend faster.
30*D(52) > 52 * D(30)
30/52 * D(52) > D(30) <=> 57% > D(30) / D(52)
Then the question boils down to if the drag at 30 tons is less than 57% of the drag at 52 tons at the given airspeed. If that is true, the heavy aircraft will descend faster.
I find it highly unlikely.
Cheers,
Fred
Doing a bit of algebra, I find that the rate of descent is proportional to the drag (D) through the mass (m) of the aircraft. The drag will of course in turn depend on the mass, D(m).
Thus, the difference in descent rate, heavy aircraft descending faster being the positive, can be described by
V/S difference = D(52)/52 - D(30)/30 = ( 30*D(52)-52*D(30) ) / (52*30)
In other words, if
30*D(52)-52*D(30) > 0, the heavy aircraft will descend faster.
30*D(52) > 52 * D(30)
30/52 * D(52) > D(30) <=> 57% > D(30) / D(52)
Then the question boils down to if the drag at 30 tons is less than 57% of the drag at 52 tons at the given airspeed. If that is true, the heavy aircraft will descend faster.
I find it highly unlikely.
Cheers,
Fred
Join Date: Apr 2004
Location: Earth
Posts: 6
Likes: 0
Received 0 Likes
on
0 Posts
Answer from a lazy pilot:
Heavy or light have same best glide distance except the lighter one fly slower IAS. To maintain same speed with heavier one, pitch down increase rate of descent.
answer: lighter a/c higher rate of descent.
Heavy or light have same best glide distance except the lighter one fly slower IAS. To maintain same speed with heavier one, pitch down increase rate of descent.
answer: lighter a/c higher rate of descent.
Join Date: May 2004
Location: prime meridian
Posts: 42
Likes: 0
Received 0 Likes
on
0 Posts
a perusal of kermode (pg192) reminds us that keith has got it nailed
having said that, a landing wt of 120 tons gives us a dist required to descend of about 10 nm more than at 100 tons at the SOP speed of .80/300 - that's where the chosen speed comes into play i guess with these speeds being well above best L/D
having said that, a landing wt of 120 tons gives us a dist required to descend of about 10 nm more than at 100 tons at the SOP speed of .80/300 - that's where the chosen speed comes into play i guess with these speeds being well above best L/D
Actually, Keith got it a bit wrong! 
In all descents at line flying speeds, the lighter aircraft will have a steeper angle, which at the same speed will mean a higher ROD.
The simple answer is that the heavier aircraft will (probably) have a lesser glide endurance and hence a greater rate of descent.
Join Date: Aug 2001
Location: Dorset
Posts: 775
Likes: 0
Received 0 Likes
on
0 Posts
Checkboard,
I think that you will find that this is because "line flying speeds" are close to Vmd and therefore higher than Vmp for both the light and heavy aircraft. This means that if they are both flying at such speeds, the heavy aircraft will be closer to its best glide endurance speed (its own Vmp) than the lighter aircraft will. So at such speeds the light aircraft will have the lesser endurance and hence higher ROD.
But if you were to glide both aircraft at the Vmp for the light aircraft it would (probably) have the greatest endurance.
FT,
Your equations are interesting but I am not convinced. If I may quote you,
"Then the question boils down to if the drag at 30 tons is less than 57% of the drag at 52 tons at the given airspeed. If that is true, the heavy aircraft will descend faster....I find it highly unlikely".
Well Cdi is proportinal to Cl squared and at a common speed the Cl at 52 tons will be 52/30 = 1.733 times that at 30 tons. So the Cdi at 52 tons will be 1.733 squared = 3.00 times sthat at 30 tons. This means that the induced drag at 53 tons will 200% greater than that at 52 tons. or to put it another way the induced drag at 30 tons is only 33% of that at 52 tons.
I think that you will find that this is because "line flying speeds" are close to Vmd and therefore higher than Vmp for both the light and heavy aircraft. This means that if they are both flying at such speeds, the heavy aircraft will be closer to its best glide endurance speed (its own Vmp) than the lighter aircraft will. So at such speeds the light aircraft will have the lesser endurance and hence higher ROD.
But if you were to glide both aircraft at the Vmp for the light aircraft it would (probably) have the greatest endurance.
FT,
Your equations are interesting but I am not convinced. If I may quote you,
"Then the question boils down to if the drag at 30 tons is less than 57% of the drag at 52 tons at the given airspeed. If that is true, the heavy aircraft will descend faster....I find it highly unlikely".
Well Cdi is proportinal to Cl squared and at a common speed the Cl at 52 tons will be 52/30 = 1.733 times that at 30 tons. So the Cdi at 52 tons will be 1.733 squared = 3.00 times sthat at 30 tons. This means that the induced drag at 53 tons will 200% greater than that at 52 tons. or to put it another way the induced drag at 30 tons is only 33% of that at 52 tons.
Join Date: Aug 2001
Location: Dorset
Posts: 775
Likes: 0
Received 0 Likes
on
0 Posts
FT,
No I did not forget it, I just did not discuss it.
At Vmd the induced drag is half of the total drag. Only the induced drag is affected by weight changes. So at Vmd the increase in total drag for any given weight increase will be half of the increase in Induced drag. In effect this means that for small % weight changes, there will be a 1 to 1 ratio between weight increase and total drag increase.
So using the figures in my previous post (Induced drag increased to 300% of its previous value). If this happens at Vmd then total drag increases to 150% of its previous value. So the drag of the 30 tonne jet is about 2/3 that of the 52 tonne jet.
But at lower speeds the induced drag is a much greater proportion of the total drag. So the increase in total drag is a much greater proportion of the increase in induced drag. The relationship between weight increases and total drag increases become smuch greater than 1 to 1. This means that at low speeds such as Vmp it is entirely possible that the drag on the 30 tonne jet will be less than 57% of that of the 52 tonne jet.
The overall effect of this drag-weight-speed relationship is that the heavy jet will have the greatest ROD at low speeds and the lighter jet will have the greatest ROD at high speeds.
Readers who prefer the "nose further down to maintain speed at low weights - therefore greater ROD" argument, should consider what happens at low speeds. The heavier aircraft will have an inefficiently high angle of attack, so it will produce a great deal of drag. This high drag will use up its energy more quickly.
The fact that gliding at or close to Vmd causes light aircraft to sink faster, does not mean that this is true at all speeds. It will just appear to be inevitable if you do most of your gliding at or close to Vmd.
No I did not forget it, I just did not discuss it.
At Vmd the induced drag is half of the total drag. Only the induced drag is affected by weight changes. So at Vmd the increase in total drag for any given weight increase will be half of the increase in Induced drag. In effect this means that for small % weight changes, there will be a 1 to 1 ratio between weight increase and total drag increase.
So using the figures in my previous post (Induced drag increased to 300% of its previous value). If this happens at Vmd then total drag increases to 150% of its previous value. So the drag of the 30 tonne jet is about 2/3 that of the 52 tonne jet.
But at lower speeds the induced drag is a much greater proportion of the total drag. So the increase in total drag is a much greater proportion of the increase in induced drag. The relationship between weight increases and total drag increases become smuch greater than 1 to 1. This means that at low speeds such as Vmp it is entirely possible that the drag on the 30 tonne jet will be less than 57% of that of the 52 tonne jet.
The overall effect of this drag-weight-speed relationship is that the heavy jet will have the greatest ROD at low speeds and the lighter jet will have the greatest ROD at high speeds.
Readers who prefer the "nose further down to maintain speed at low weights - therefore greater ROD" argument, should consider what happens at low speeds. The heavier aircraft will have an inefficiently high angle of attack, so it will produce a great deal of drag. This high drag will use up its energy more quickly.
The fact that gliding at or close to Vmd causes light aircraft to sink faster, does not mean that this is true at all speeds. It will just appear to be inevitable if you do most of your gliding at or close to Vmd.
Last edited by Keith.Williams.; 13th May 2004 at 06:47.
Join Date: Jun 2003
Location: Shoreline
Posts: 39
Likes: 0
Received 0 Likes
on
0 Posts
In my experience:
95,000lbs, 250 kts = 1,500 fpm
78,000lbs, 250 kts = 1,800 fpm
my answer? light aircraft, faster descent, same forward velocity.
But I do appreciate this discussion! Very nice!
I only wish the question should have been more specific.
95,000lbs, 250 kts = 1,500 fpm
78,000lbs, 250 kts = 1,800 fpm
my answer? light aircraft, faster descent, same forward velocity.
But I do appreciate this discussion! Very nice!
I only wish the question should have been more specific.
Thread Starter
Join Date: Oct 2003
Location: Lee Gardens
Posts: 45
Likes: 0
Received 0 Likes
on
0 Posts
wow, shiver me timbers
i am glad to have brought this up again cuz' i know zilch.
yes, it was hashed some time back and KE=1/2 mv(squared) was used to illustrate this.
it was proven (altough with much debate) that a lighter a/c would have a higher ROD than a heavier a/c.
sorry i have not put in my response, blasted thing went dead on me.
yes, it is about rate of descent, two perfectly fine aeroplanes descending at the same speed. with thrust at idle.
yes, it was hashed some time back and KE=1/2 mv(squared) was used to illustrate this.
it was proven (altough with much debate) that a lighter a/c would have a higher ROD than a heavier a/c.
sorry i have not put in my response, blasted thing went dead on me.
yes, it is about rate of descent, two perfectly fine aeroplanes descending at the same speed. with thrust at idle.
