onenastyviper
|
 posted on 2/9/11 at 09:24 AM |
|
|
Dynamic Weight Transfer
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?
|
|
|
|
|
britishtrident
|
| posted on 2/9/11 at 10:23 AM |
|
|
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.
|
|
|
Sam_68
|
| posted on 2/9/11 at 11:26 AM |
|
|
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.
|
|
|
onenastyviper
|
| posted on 2/9/11 at 12:18 PM |
|
|
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.
But doesn't the height of the CG play an important part in weight transfer?
All the equations I have found illustrate this and the equation I have derived also illustrates it:
TotalDynamicFrontWeight = StaticFrontWeight + DynamicFrontWeight
From 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.
http://www.sae.org/students/presentations/brakes.ppt
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)?
[Edited on 2/9/11 by onenastyviper]
|
|
|
britishtrident
|
| posted on 2/9/11 at 01:40 PM |
|
|
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.
[I] “ What use our work, Bennet, if we cannot care for those we love? .”
― From BBC TV/Amazon's Ripper Street.
[/I]
|
|
|
v8kid
|
| posted on 2/9/11 at 02:58 PM |
|
|
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
You'd be surprised how quickly the sales people at B&Q try and assist you after ignoring you for the past 15 minutes when you try and start a
chainsaw
|
|
|
Sam_68
|
| posted on 2/9/11 at 04:36 PM |
|
|
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)?
I'd say you've also definitely got your weight distribution wrong... 74% F 26% rear, with driver?!
...no wonder it's standing on its nose when (in virtual reality) you brake!
60% F/40% R is more like it, and that's without driver.
Your front axle mass can exceed the total vehicle weight in calculation but as V8kid says, it means that your rear axle weight is
negative, therefore if it happened in reality, the car would be somersaulting end-over-end.
Just as if you do the same calculation for lateral weight transfer and get a negative inside weight/more-than-100% outside weight, it means
(mathematically) your car has rolled on the bend.
[Edited on 2/9/11 by Sam_68]
|
|
|
ettore bugatti
|
| posted on 2/9/11 at 06:24 PM |
|
|
CoG should be nearer 0.55-0.65m of ground.
|
|
|
