- 47
- Glasgow
I was reading gt5mania.com and thought I would try out the Weight transfer formula in GT5 on the BMW M5, but I would like help confirming where the M5 centre of gravity is.
Length: 4855mm
Height: 1469mm
Width: 1846mm
Centre of Gravity=(Length1*Weight1)+(Length2*Weigh t2)/Total Weight
As I only have the overall Weight & Length, the CoG would be the same as the length, so with that in mind, I have set CoG at half length of the entire car, and if the weight is evenly dispersed then I think it will be a fair value.
mass=1855kg
Weight=massxgravitational pull=1855*9.81=18197.55N
Force=massxacceleration=1855kg*62.5mph(1 00.5625kmh)=186.543MN
accleration=finalvelocity-initialvelocity/time=60mph(96.54kmh)-0mph/5sec=11.758mph/sec (19308m/s/s OR 19.308kmh)
deceleration=60mph (96.54kmh)/4sec=14.698mph/sec (24135m/s/s OR 24.135kmh) (.668G)
1G=32f/s (22mph)
(Front Weight Transfer) Force Down = Braking 'G' x mass x height/a
It appears that "a" is pitch centre, of front part of vehicle. In the example on GTMania pitch centre appears to be determinable by Braking G Force*Vehicle Centre of Gravity. However, I expect the relativity to change according to vehicle, so a way of determining BMW M5 pitch centre will also be helpful.
http://gt5mania.com/driving-guide/st...ight-transfer/
Front Pitch "a" = Braking G Force * Vehicle CoG = .668*2427.5 = 1621.57N
Rear Pitch "b" = Braking G Force * Vehicle CoG = .668*-2427.5 = -1621.57N
With the statements of CoG and pitch in mind, lets now try the formula:
(Front Weight Transfer) Force Down = Braking 'G' x mass x height/a
= 0.668G*1855*4855/1621.57= 3710N= 378.42kg/f (GAINED)
(Rear Weight Transfer) Force Up = Braking 'G' x mass x height/b
=.668G*1855*4855/1621.57=3710N=378.42kg/f (LOST)
Remember, I am assuming the Weight, Mass are evenly dispersed. Therefore;
Initial Front, Rear Mass = 927.5kg
Initial+Transfer = 927.5+378.42=1305.92kg (Front Mass After Weight Transfer)
Initial -Transfer=927.5-378.42=549.08kg (Rear Mass After Weight Transfer)
Total Mass = 1305.92 + 549.08 = 1855kg
/Thanks/
Length: 4855mm
Height: 1469mm
Width: 1846mm
Centre of Gravity=(Length1*Weight1)+(Length2*Weigh t2)/Total Weight
As I only have the overall Weight & Length, the CoG would be the same as the length, so with that in mind, I have set CoG at half length of the entire car, and if the weight is evenly dispersed then I think it will be a fair value.
mass=1855kg
Weight=massxgravitational pull=1855*9.81=18197.55N
Force=massxacceleration=1855kg*62.5mph(1 00.5625kmh)=186.543MN
accleration=finalvelocity-initialvelocity/time=60mph(96.54kmh)-0mph/5sec=11.758mph/sec (19308m/s/s OR 19.308kmh)
deceleration=60mph (96.54kmh)/4sec=14.698mph/sec (24135m/s/s OR 24.135kmh) (.668G)
1G=32f/s (22mph)
(Front Weight Transfer) Force Down = Braking 'G' x mass x height/a
It appears that "a" is pitch centre, of front part of vehicle. In the example on GTMania pitch centre appears to be determinable by Braking G Force*Vehicle Centre of Gravity. However, I expect the relativity to change according to vehicle, so a way of determining BMW M5 pitch centre will also be helpful.
http://gt5mania.com/driving-guide/st...ight-transfer/
Front Pitch "a" = Braking G Force * Vehicle CoG = .668*2427.5 = 1621.57N
Rear Pitch "b" = Braking G Force * Vehicle CoG = .668*-2427.5 = -1621.57N
With the statements of CoG and pitch in mind, lets now try the formula:
(Front Weight Transfer) Force Down = Braking 'G' x mass x height/a
= 0.668G*1855*4855/1621.57= 3710N= 378.42kg/f (GAINED)
(Rear Weight Transfer) Force Up = Braking 'G' x mass x height/b
=.668G*1855*4855/1621.57=3710N=378.42kg/f (LOST)
Remember, I am assuming the Weight, Mass are evenly dispersed. Therefore;
Initial Front, Rear Mass = 927.5kg
Initial+Transfer = 927.5+378.42=1305.92kg (Front Mass After Weight Transfer)
Initial -Transfer=927.5-378.42=549.08kg (Rear Mass After Weight Transfer)
Total Mass = 1305.92 + 549.08 = 1855kg
/Thanks/
Last edited: