DBK
LIFE MEMBER
I'm not sure why this thread has been exhumed but rereading it I'm sorry but this explanation is wrong, sorry I missed it the first time around.The vast majority of replies seems hooked on "Load" and sharing the weight.
Now assuming for one moment DBK.s and the rest are all 100% correct it still doesn't answer the question asked
No one has asked what that load will be, or which axle will carry what , shared or otherwise
Please answer the question
For the purpose of the Formula and Calculations ONLY ------Which axle is used ? (see diagram) and the same question asked by Jim in the original post.
Use anything other than the center of the rear most axle such as the center of the two or the front tandem axle and you will in effect be incorrectly CALCULATING The wrong overhang figure
Over simplified BUT place a 100 Ton down force on the tow ball and the vehicle WILL swivel on its rearmost wheels, and its the distance from the swivel point and the tow bar that is needed for the calculation regardless of what weight it turns out to be or whether its shared or not
As a downward force is applied to the towbar the effect is to gradually move the centre of gravity rearwards as the force increases. If you first imagine a vehicle with solid wheels and no suspension nothing visible would happen at first but when the force became large enough (in reality several times the weight of the actual vehicle) a point would be reached when the centre of gravity would move past the rear tag axle and the vehicle would then rotate around the rearmost axle and both the front and forward tag axle would lift off the ground.
In reality of course MHs don't have solid wheels and rigid suspension so what happens is as a force is applied downwards on the towbar the rear of the vehicle sinks (both tag axle springs compressed) and the front of the vehicle lifts slightly. Seen from the side the vehicle would rotate around a point somewhere between front and rear axles. It isn't possible to calculate where this point is because it would depend on the spring rates for each axle. To take an extreme case if the front axle had a spring so hard it was effectively rigid then the vehicle would rotate around the front axle, bizarre as that might sound.
I think for all practical purposes if calculations were based on assuming the vehicle had a single rear axle mounted mid-way between the tag axles the figures obtained would be good enough, taking into account the other vagaries of loading such a water, fuel and waste etc. The extra loading calculated on this imaginary single axle could be shared equally between the tag axles, that is 50% each.
The rearmost axle will actually carry a little more load than the forward tag but I doubt there would ever be more than 10Kg or at most 20Kg difference between the axles if the extra load was something like a scooter. If things were this close to the limit then the viability of the extra load (scooter) would have to be questioned.
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