Vmbe
 

I am trying to reconcile the actual with the theoretical and having trouble with it - possibly the units.

Vmbe is a function of how much kinetic energy the brakes can stop (using worn brakes to the overhaul limit).

So with 70mj brakes the kinetic energy is = 1/2mv^2.

If you know your mass (as you would for take off!) you should be able to get a reasonable approximation of V1?

However, I have tried all sorts of permutations with units but cannot make it work, even roughly. I have tried to reverse the formula using performance data but cannot get anything within the range of 70mj (in order to account for the worn brakes).

Can anyone help?

Depends on the aircraft of course, but a 50 tonne aircraft (50,000kg;110,000lb) with a V1 of 100knots (51.4 m/s) would have a kinetic energy at V1 of:

0.5 * 50,0000 * 51.4 * 51.4 = 66.049 MJ which is certainly in the range you want.

Things to look out for are:
the speed is true ground speed, not airspeed, so you need to correct for wind and altitude effects
Vmbe may assume some energy goes into drag or TR use (depending on whether the energy quoted is the BRAKE capacity or the aircraft energy at which the brakes hit the limit)

hmmm
 

Thanks MFS,

Here are some actual figures from a perfomance program after a Vmbe calc.
70mj brakes, MTOW 75000kg, V1max 170kts, Brake Energy 100%.
so...5*75000*(84.5^2) equals 267759375 or 26mj...well off!

Dry runway, sea level, no wind, no THR REV.

Err, small arithmetical mistake there...

0.5*75000*(84.5^2) equals 267,759,375 or 268MJ

70*4=280 - that "70MJ" brake rating wouldn't happen to be "per brake" by any chance? If so, it's pretty close for a four brake case.

Yes, never thought of the per brake issue - and there are four brakes on the aircraft!

I should think that solves it - I will run some more cases and see that it roughly works.

Thanks!

CS