Good morning everyone, I know it is early so I will be gentle 
I am playing about with the equations for dynamic weight distribution front to rear (2D only).
My question is this, I plugged some numbers in for my fiesta and under a 1g brake decceleration, the front axle is higher than the total mass of the
vehicle, is this dynamically possible?
I am thinking that if the front axle dynamic mass is higher than the total mass then something bad is happening at the rear end and so the maximum
braking decceleration is limited to maximum weight on front axle?
Your maths are wrong somewhere, good old fashioned principle of moments about the front wheel contact point. If there is no aero downforce
or lift the sum of vertical component of forces s at supports will always equal Mg. As a rull of thumb In a small short wheelbase FWD hatch
back almost but not quite 100% of the braking is done by the front wheels.
For a 1g stop with 100% weight transfer for a car weighing 1000kg the horizontal deceleration force will be roughly 10kn the vertical
weight component will be roughly 10kn and the reesultant force through the suspension is roughly 1.42 kn at 45 degrees from the horizontal acting
through the front tyre contact patches.
Yes, it's dynamically/mathematically possible - it's what would be happening if you did a 'stoppy' on a motorcycle and failed to
ease off the brake when the rear tyre lost contact with the ground... basically the bike would somersault forwards over the front wheel.
If you're getting that result with your calculation, it's telling you that all the weight would have been transferred onto the front wheels
and the rear tyres would be off the ground somewhere before you reach 1g deceleration.
Common sense suggests that this is unlikely, however, so you've probably got something wrong with your sums. My best guess, assuming that
the actual calculation methods are correct, is that you're estimating the height of your centre of gravity too high.
quote:
Originally posted by britishtrident
Your maths are wrong somewhere, good old fashioned principle of moments about the front wheel contact point. If there is no aero downforce or lift the sum of vertical component of forces s at supports will always equal Mg. As a rull of thumb In a small short wheelbase FWD hatch back almost but not quite 100% of the braking is done by the front wheels.
For a 1g stop with 100% weight transfer for a car weighing 1000kg the horizontal deceleration force will be roughly 10kn the vertical weight component will be roughly 10kn and the reesultant force through the suspension is roughly 1.42 kn at 45 degrees from the horizontal acting through the front tyre contact patches.
You can only have 100% transfer because total of vertical reactions = down force due to weight. If you have more than that you the system becomes dynamic resulting in an end over end roll.
You have negative weight on the rear axle which causes upward acceleration in opposition to gravity. Modify your spreadsheet to accept only positive
weights and all will be OK.
Cheers1
quote:
Originally posted by onenastyviperFrom my car:
Ford Fiesta
2.5m wheelbase
700kg on the front axle static
250kg rear axle static
and cg height estimated at 0.75m from ground level
For a 1g decceleration, I calculated approximately 985kg on the front axle >950kg total car mass.
...Of course I could have got the CG height wrong but my question is can the front axle mass ever exceed the total vehicle mass under braking (or acceleration)?
CoG should be nearer 0.55-0.65m of ground.