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Hachiouji-shi
18th Mar 2006, 15:27
Why do we reduce the wet V1 lesser and lesser with increasing aircraft take off weight?

one dot right
18th Mar 2006, 15:33
Er, cos it'll take longer and longer to stop?:eek:

alexban
19th Mar 2006, 18:49
increasing weight---increasing braking force

Hachiouji-shi
20th Mar 2006, 11:59
shouldn't the V1 be decreased more if the aircraft is heavier because it has more momentum and hence more distance required to stop?

an excerpt from the 737 QRH:

Weight V1 Adjustments

68000 minus 4 kts

60000 minus 7

52000 minus 10

44000 minus 12


Why ?

machone
20th Mar 2006, 13:29
Manufacturers usually pick a mass to use as the reference point. There after all weights and speeds are plus or minus. So as the mass goes down you must subtract from the reference speed if the mass were to go up you would add

Hachiouji-shi
20th Mar 2006, 14:10
I dont get it.

captainpaddy
20th Mar 2006, 14:41
I don't really get it either. However, the distance covered at 68T while accelerating by 4 knots would be much greater than that covered while accelerating by 4 knots at 44T. Perhaps this is why there is a larger speed reduction at the lower weights? The 12 knots at 44T results in a more significant distance reduction than 4 knots (or less) would because you are accelerating much quicker. At high weights you cover a large distance for every extra knot because your acceleration is so poor. My mind works in terms of distance when it comes to corrections, so ignore the speed value and think of the distance recovered by the reduction?

Don't know if that helps!!

CP

Hachiouji-shi
20th Mar 2006, 14:49
I am almost getting there....

or so I thought... :(

captainpaddy
20th Mar 2006, 15:06
How about:..........

At high weight:
A very large distance will be required to accelerate and then stop if you have a problem. If it's wet, the reduction in V1 allows enough room to stop (with less efficiency cos it's wet) by forcing the abort decision to be made earlier (by reducing the V1 speed). 4 knots of acceleration at a high weight equals a lot of tarmac, cos it's takes a longer time achieve. So, a significant distance is recovered for the stop case.

(Totally made up figures)
Weight 68T dry balanced field 8000 feet long
Distance required to reach V1 at 150 knots: 5000 feet
Distance left to stop from V1: 3000 feet

Weight 68T same runway but WET
Distance required to reach new V1 at 146 knots: 4500 feet
Distance left to stop from V1: 3500 feet (500 feet extra room for reduced stopping performance)

Now,
Weight 44T, balanced field 5500 feet long
Distance required to reach V1 at 125 knots: 3000 feet
Distance left to stop from V1: 2500 feet

Weight 44T same runway but WET
Distance required to reach new V1 of 113 knots: 2600 feet
Distance left to stop from V1: 2900 feet (400 feet extra room for reduced stopping performance. Less than we recovered at the heavy weight but since we're lighter and at aborting from a lower speed it will be enough)

Imagine if we only reduced the wet V1 in the light weight case by 4knots. Because we are accelerating so quickly we would reach the new V1 (121 knots) in say 2900 feet, which would only give us an extra 100 feet for stopping. So the larger speed reduction recovers more tarmac!

Remember to think of the high and low weight cases as completely seperate cases on completely different runways. The light weight scenario would not exist on the same runway as the heavy one, because you wouldn't be limited by the stopping distance at all, regardless of your V1 at a light weight, because the runway would have to be so huge for the heavy weight case to get off the ground. (That's why I've used different made up balanced field lengths above)

???