Vmcg Limit Weight
Thread Starter
Join Date: Oct 2005
Location: Away
Posts: 300
Likes: 0
Received 0 Likes
on
0 Posts
Vmcg Limit Weight
Hi there.
I wonder if someone knowledgeable on performance matters might shed some light on why we use Vmcg Limit Weight, and why it increases with an increasing depth of runway contamination.
For a B744 Slush/Standing Water Takeoff on an 11,400' runway the Vmcg Limit Weight is 381 tonnes in 3mm of (uniform) contamination. The same plane on the same runway length has a Vmcg Limit Weight of 402 tonnes in 6mm of contamination.
I recall the Performance fellow from class mentioning that deeper contamination has a greater slowing effect on the aircraft during a rejected takeoff. This made sense as there's more goo to slow the plane down. Consequently, a higher Vmcg Limit Weight is achieved with increasing contamination, up to the fuselage damage-protecting limit of 13mm.
But I doubt my recollection, as Vmcg has to do with controllability, not stopping capability. That's a V1 calculation.
Does the equation Speed = Distance divided by Time come into this?
To achieve an all-but-constant Vmcg speed takes more time, and distance, if wading through heavier contamination (friction) than a less contaminated surface. Easy enough.
So why would the Vmcg Limit Weight increase with an increasing amount of contamination? You've got more friction for the heavier plane to wade through. It'll therefore get to its Vmcg speed at a greater distance along the runway. At some point, the plane will have not enough distance to stop. The point at which the Vmcg and V1 speed are reached at the same time, therefore distance, assuming all other factors are equal, must be the Vmcg Limit Weight.
Is that right?
I'm sure I have it all confused.
I wonder if someone knowledgeable on performance matters might shed some light on why we use Vmcg Limit Weight, and why it increases with an increasing depth of runway contamination.
For a B744 Slush/Standing Water Takeoff on an 11,400' runway the Vmcg Limit Weight is 381 tonnes in 3mm of (uniform) contamination. The same plane on the same runway length has a Vmcg Limit Weight of 402 tonnes in 6mm of contamination.
I recall the Performance fellow from class mentioning that deeper contamination has a greater slowing effect on the aircraft during a rejected takeoff. This made sense as there's more goo to slow the plane down. Consequently, a higher Vmcg Limit Weight is achieved with increasing contamination, up to the fuselage damage-protecting limit of 13mm.
But I doubt my recollection, as Vmcg has to do with controllability, not stopping capability. That's a V1 calculation.
Does the equation Speed = Distance divided by Time come into this?
To achieve an all-but-constant Vmcg speed takes more time, and distance, if wading through heavier contamination (friction) than a less contaminated surface. Easy enough.
So why would the Vmcg Limit Weight increase with an increasing amount of contamination? You've got more friction for the heavier plane to wade through. It'll therefore get to its Vmcg speed at a greater distance along the runway. At some point, the plane will have not enough distance to stop. The point at which the Vmcg and V1 speed are reached at the same time, therefore distance, assuming all other factors are equal, must be the Vmcg Limit Weight.
Is that right?
I'm sure I have it all confused.
Join Date: Sep 1999
Location: ME
Posts: 5,502
Likes: 0
Received 0 Likes
on
0 Posts
The chart that you are looking at is a V1=V1MCG limit chart, therefore V1 does play a role in the weight calculation.
For a greater depth of contaminant you will get a higher takeoff weight due to the "ploughing" effect encountered during the stop, this aids the aircraft in stopping and allows the V1 to be increased, the higher V1 permits a higher limiting weight.
If you are still in doubt, ask your training people to show you the Boeing presetation on the subject.
Mutt
For a greater depth of contaminant you will get a higher takeoff weight due to the "ploughing" effect encountered during the stop, this aids the aircraft in stopping and allows the V1 to be increased, the higher V1 permits a higher limiting weight.
If you are still in doubt, ask your training people to show you the Boeing presetation on the subject.
Mutt
Join Date: Jan 2006
Location: Airplane
Posts: 133
Likes: 0
Received 0 Likes
on
0 Posts
I can't say this is 100% correct but it seems to me that the more contaminated a runway is the more slipery it will be. If this is true then the traction would also be less when it comes to controling an engine out scenerio. Because if this the rudder must be able to compensate and this is done by giving you a higher speed, hence a higher weight. Of course this only comes into play if there are no other limiting performance factors.
7
7
Join Date: May 1999
Location: Bristol, England
Age: 65
Posts: 1,806
Received 0 Likes
on
0 Posts
Not sure about your second paragraph, Mutt. Looking at the B737 graphs I have for Mass and V1 reductions you seem to get a lower take-off weight as depth of contaminant increases and a higher V1.
I interpret this as braking performance being fairly equally poor whether you have 2mm or 13mm of slush, and therefore the VSTOP vs weight line stays fairly steady, but the impingement drag caused by the greater contaminant depth causing the minimum speed to go, VGO, to increase at any given weight. The effect of a relatively stationary VSTOP line and VGO increasing is to reduce MTOM and increase V1.
Am I using these incorrectly? I'd appreciate the advice.
I interpret this as braking performance being fairly equally poor whether you have 2mm or 13mm of slush, and therefore the VSTOP vs weight line stays fairly steady, but the impingement drag caused by the greater contaminant depth causing the minimum speed to go, VGO, to increase at any given weight. The effect of a relatively stationary VSTOP line and VGO increasing is to reduce MTOM and increase V1.
Am I using these incorrectly? I'd appreciate the advice.
Join Date: Sep 1999
Location: ME
Posts: 5,502
Likes: 0
Received 0 Likes
on
0 Posts
airbus757, you are presuming that VMCG is calculated for contaminated runways. Certainly this isnt the case for FAR's, we only have 1 set of values for VMCG as determined in FAR25.
Alex, looking at the presentation that i mentioned, it shows the same trend for B737-500 aircraft, willing to send you the presentation if you let me know your email.
Cheers
Mutt
Alex, looking at the presentation that i mentioned, it shows the same trend for B737-500 aircraft, willing to send you the presentation if you let me know your email.
Cheers
Mutt
Join Date: Aug 2001
Location: Dorset
Posts: 775
Likes: 0
Received 0 Likes
on
0 Posts
Alex,
If you are refering to the CAP 698 then I suspect that it is a bit more complicated than simply reducing weight and increasing V1 as contamination depth increases.
The main tables on pages 72 and 73 certainly show both TOM and V1 decreasing as mass increases.
But the VMCG Limited Mass tables show that the mass increases as contamination depth increases. At 6600 ft field length and 4000 ft PAlt for example the figures are 45K at 2 mm depth, 48K at 6mm and 52K ar 13 mm depth.
If you are refering to the CAP 698 then I suspect that it is a bit more complicated than simply reducing weight and increasing V1 as contamination depth increases.
The main tables on pages 72 and 73 certainly show both TOM and V1 decreasing as mass increases.
But the VMCG Limited Mass tables show that the mass increases as contamination depth increases. At 6600 ft field length and 4000 ft PAlt for example the figures are 45K at 2 mm depth, 48K at 6mm and 52K ar 13 mm depth.
