Optimum Acceleration and Braking Force Distribution Curve - part 2 with EXCEL Program

Contents

1. Sign convention on the direction of acceleration and braking force
2. Condition of the optimum acceleration and braking force distribution curve
3. Relationship between tire friction force and gravitational acceleration
4. Basic equation for the optimum force distribution curve
5. Practical example for a car
6. Analysis of the optimum force distributio curve
7. Theoretical base for the important values of curve
8. Conclusion

sign convention on the direction of acceleration and braking force

Sign convention is redefined to explain both of acceleration and braking

Plus sign is for acceleration

Minus sign is for braking

The shape of the brake force distribution graph moves from upper-right quadrant to lower-left quadrant. it is a point symmetry of the previous one as shown as following graph.

Longitudinal force distribution(a) - optimum case

Explained in the previous article, the vehicle weight can be expressed as two equations here 

in which a is plus for acceleration and minus in braking. 

Front weight decreases in acceleration and increases in braking.

From the moment equation about y axis at rear tire-road contact center,

In the optimum case of longitudinal force distribution, all the tire friction coefficients "µ" are the same at all 4 wheels  and in addition to that, all 4 tires reach their maximum forces at the same time. There is no need to worry about any yawing moment except for the laterally big offset of CG.

Newton’s law says that Sum of reaction forces should be equal to the vehicle inertia force due to acceleration. So, we have equation A③

substituting A① and A② for Fzf +Fzr of  A③, we can end up with the simple equation MEU is the ratio of vehicle acceleration a to gravitational acceleration g

Equation of optimum longitudinal force distribution

Assuming that µ is the same value at all 4 wheels and all tires reach their maximum forces at the same time,

substituting these two equations for Fzf and Fzr of A① and A② and employing µ = a/g,

Dividing B③ and B④ by vehicle weight mg to make the longitudinal forces dimensionless,

Arranging the equations

B⑦ and B⑧ are the final equations to draw the optimum longitudinal force distribution curve. These two equations here are the quadratic functions of the ratio of the vehicle acceleration to the gravitational acceleration.

presented above, there is no term of vehicle weight in the right side of equations. they have only geometric data and the ratio of the vehicle acceleration to the gravitational acceleration.

EXCEL sheet for optimum longtudinal force distribution

All we need are geometry data for optimum longitudinal force distribution curve. No data is required for the weight.

In the red rectangular, we have an example of the car data set including the geometry of CG location. 
Step 2 is for calculating 2 values of 3.86 and -2.57. what are those 2 values here?  3.86 is the g-units for front tire to lose the ground contact and called ‘tilting acceleration’ and -2.57 is the g-units for rear tire to lose the ground contact and called ‘tilting deceleration’ we are going to draw the optimal longitudinal force distribution curve between those two ratio values of a to g.

Example of Acc. And Brake Distribution curve with EXCEL

As a result, we can draw the braking and the acceleration force distribution curves. 

The curve in color of royal blue is for front wheels and the other in the color of khaki is for rear wheels. 

 We have tilting deceleration at which the slope is half a ratio of L sub f to wheelbase L, the value of which is -0.2 here and tilting acceleration at which the slope is half a ratio of L sub r to wheelbase L, value of which is -0.3 here

The normalized maximum force at the rear wheel in the braking reaches its maximum at half a ratio of L sub f to CG height h the value of which is -1.29 here

 and the normalized maximum force at the front wheel in the acceleration reaches its maximum at half a ratio of L sub r to CG height, the value of which is 1.93 here. How do I know that? I will make that clear soon.

Maximum forces for braking and acceleraton

Maximum force at front wheel in the acceleration can be obtained from the derivative of the equation 𝑩⑦.

At the maximum, derivative of 𝐹𝑥𝑓/𝑚𝑔 with respect to 𝑎/𝑔 should be zero.

Maximum force at rear wheel in the Braking can be obtained from the derivative of the equation B⑧.

At the maximum, derivative of 𝐹𝑥𝑟/𝑚𝑔 with respect to 𝑎/𝑔 should be zero.

substituting those two values for a/g in 𝑩⑦ and  B⑧, maximum forces can be obtained.

Slope at 𝐹𝑥𝑓/𝑚𝑔=0,  𝐹𝑥𝑟/𝑚𝑔=0

Slope at 𝐹𝑥𝑓/𝑚𝑔=0 in the acceleration can be obtained from the derivative of the equation 𝑩⑦.

From the equation 𝑩⑦, the roots of 𝐹𝑥𝑓/𝑚𝑔 are 𝑎/𝑔=0, L𝑟/ℎ.

 

The absolute values are the same.

Slope at 𝐹𝑥𝑟/𝑚𝑔=0 in the Braking can be obtained from the derivative of the equation B ⑧.

From the equation B ⑧, the roots of 𝐹𝑥𝑟/𝑚𝑔 are 𝑎/𝑔=0, −  L𝑓/ℎ.

The absolute values are the same.

Conclusion

1. The longitudinal force distribution curve is exclusively determined by the geometric data of the vehicle.
2. Maximum acceleration forces of the front wheels are the function of CG location
3. Maximum braking force of the rear wheels are also the function of CG location
4. With the enough powertrain capacity, the minimum time for given distance relies on tire friction force.
5. Tilting acceleration and tilting deceleration is inversely proportional to the CG height and is proportional to the distance from CG to the front axle and the rear axle respectively
6. Maximum acceleration force at front wheel and maximum braking force at rear wheel are inversely proportional to the CG height and is proportional to the distance form CG to the front axle and to the rear axle respectively