Have finally been driven to pull out my '72 edition of Kermode's
Mechanics of Flight. Appendix 2 presents acceleration formulae in various units.
Taking
4b. F = (W/g)a:
a 744 at 644,000 lbs with autobrakes producing a 10 ft/s/s deceleration and g of 32.2 ft/s/s produces a braking force of:
F = (644,000/32.2)10 = 200,000 pounds
Wildly assuming a CG 15 ft above the ground for this tall a/c (pending a better figure from the more knowledgeable) produces a nose-down moment of 3,000,000 ft-lb.
Assuming a mlg to nosewheel distance of 100 ft. produces a corresponding nosewheel loading of 30,000 (3,000,000/100) lbs from braking.
Assuming a static nosewheel loading of some 5% produces some 30,000 lbs of loading from the CG before the mains.
So braking near autobrake max potentially
doubles nosewheel loading.
Poor braking action can drastically reduce the extra loading on the nosewheel, but there remains a potential that as deceleration increases as a result of some backstick, greater amounts can be progressively applied.
The maximum case is nosewheel loading reduced to zero by back stick with good braking action. The mlg loading will in this case be increased by:
- the static and dynamic loading previously on the nose wheel
- the tailplane downforce (approximately equal to the nosewheel loading)
which comes to some 120,000 pounds, i.e. some 20% increase in mlg loading.
I would welcome the application of precise numbers for any type.