POWERTRAIN - POSITIONING

Question 1

How do you calculate static rear axle load W_R from CoG position?

Answer 1

W_R = W × (l_m / L) where l_m is the CoG distance from the front axle and L is wheelbase.

Question 2

How do you calculate static front axle load W_F?

Answer 2

W_F = W − W_R.

Question 3

What is the friction coefficient μ definition used in the lecture?

Answer 3

μ = F / W, where F is friction force and W is vertical load.

Question 4

How does the friction concept apply to tyres in acceleration/braking?

Answer 4

Only a certain amount of torque (either accelerating or braking) can be applied before tyre grip is lost and the vehicle becomes out of control.

Question 5

In simple tyre-grip terms, what sets the maximum longitudinal force at an axle?

Answer 5

Approximately F_max ≈ μ · W_axle (using μ = F/W applied to tyre normal load).

Question 6

What is longitudinal load transfer?

Answer 6

A change in front/rear axle loads caused by acceleration or braking due to forces acting at different heights, creating a couple (moment).

Question 7

Under acceleration, which axle gains load and what happens visually?

Answer 7

Rear axle load increases (+ΔW_Y); front axle load decreases (−ΔW_Y); the front lifts and rear squats.

Question 8

Under braking, which axle gains load and what happens visually?

Answer 8

Front axle load increases (+ΔW_Y); rear axle load decreases (−ΔW_Y); the front dives and rear lifts.

Question 9

What is the Braking Axle Loads via Load Transfer Equation

Answer 9

ΔW_Y = ±(Fh/L) = (Wμh/L).

Question 10

What do F, h, L, W, μ represent in ΔW_Y = ±(Fh/L)?

Answer 10

F = braking/traction force [N]; h = CoG height [m]; L = wheelbase [m]; W = total axle load (can include downforce) [N]; μ = tyre friction coefficient.

Question 11

Why does a low CoG and long wheelbase reduce load transfer?

Answer 11

Because ΔW_Y ∝ (h/L); smaller h and larger L reduce ΔW_Y.

Question 12

Why is ~50:50 weight distribution often argued as “optimum” for handling?

Answer 12

It can maximise tyre contact utilisation under varying circuit/road conditions (balanced axle loading).

Question 13

BRUH

Answer 13

UR MUM

Question 14

Why do single-seater race cars often target ~45:55 front:rear weight distribution and wider rear tyres?

Answer 14

A rear bias can help traction/performance, and wider rear tyres help tolerate higher rear loads; ballast/auxiliaries are packaged to hit the target distribution.

Question 15

Where should variable mass (fuel/oil) ideally be located in motorsport, and why?

Answer 15

As near as possible to the CoG to avoid changing handling balance as mass changes during a race.

Question 16

Why are many road cars historically front powertrain layout?

Answer 16

Typically cheaper and simpler packaging (e.g., no propshaft requirement for many layouts), and common industry practice.

Question 17

Why do small high-performance FWD cars often have a front-heavy weight bias?

Answer 17

Tighter packaging constraints make an unfavourable bias towards the front usually unavoidable.

Question 18

Why can rear bias help “ultimate braking performance” in supercars (per the lecture)?

Answer 18

Rear bias helps offset forward weight transfer under braking; wider rear tyres help withstand the loads.

Question 19

What is the EV “skateboard” chassis concept and why does it help weight distribution?

Answer 19

The battery tray is integrated into the floor pan; it improves cabin space and lowers CoG, improving handling despite larger mass.

Question 20

What is the lecture’s key conclusion about powertrain positioning and handling?

Answer 20

Powertrain positioning strongly influences CoG location and therefore load transfer during acceleration and braking; EVs can distribute mass better and place batteries low to improve CoG and handling.