Ziggy:..'g' is the gravitational acceleration 9.8 mps.It's a fixed number.
Empty cruise: let's put it this way:
we have two TOW's -48 and 53T.
The available braking force at M=48T is lower than the available braking force at M=53T (F=M*g*mju g=9.8,mju =braking coeficient,the same in this case )
Now we accelerate both planes. At V1(48T) M48 will stop safely (available braking force=required braking force) ,M53 will also stop safely (available braking force>required braking force V1(48)<v1(53) )
We continue to accelerate:
At V1 53T (more than 10 kts higher in this case):
The M53 will stop safely -it's calculated that the available braking force=required braking force to stop the plane on the rwy
But in the M48 case, the available braking force it's lower than the M53 case (lower mass) .Indeed ,also the required braking force it's also lower due to the same reason.
But will the available braking force be higher then the required braking force? Why then the lower V1(48T)? My guess,maybe due to lower Vr and V2,optimum climb,etc...but it's a guess.
I think Old Smokey is right,but I would prefer to have some calculations done ,or the AFM saying this,not just some smart dispatcher(at our company).It says nothing regarding this subject on our AFM ,or FCTM.
Actually Boeing performance is giving us graphs where you use the assumed temperature you've found from the Airport Analysis ,the field elevation and ACTUAL TOW to find the take off speeds. But some smart guy is thinking now,that we should not use this tables,just the airport analysis,as I've told you.
One funny thing,which I've seen in real life:
Which will stop faster :a 70T tank or a nice Mercedes (with ABS,EPS...),both running at 80Kmph ?
You can bet on the tank,it's amazing.It is also on Discovery Channel,some documentary regarding new weapons.
Of course,different mass,also different mju,but nice to see ,though.
Brgds....
Alex