4PW's
21st Mar 2006, 09:07
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.