The image represents reaction at the outer wheel of a vehicle moving along a level circular path.

To compute for the reaction, six essential parameters are needed and these parameters are **Mass of the Vechicle (m), Acceleration due to Gravity (g), Velocity of the Vehicle (v), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).**

The formula for calculating the reaction at the outer wheel of a vehicle moving along a level circular path:

R_{B} = ^{mg} / _{2}[1 + ^{v²h}/_{gra}]

Where;

R_{B} = Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path

m = Mass of the Vechicle

g = Acceleration due to Gravity

v = Velocity of the Vehicle

h = Height of Centre of Gravity of the Vehicle from Ground Level

r = Radius of Circular Path

a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;

Find the reaction when Mass of the Vechicle (m) is 12, Acceleration due to Gravity (g) is 9.8, Velocity of the Vehicle (v) is 28, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 16, Radius of Circular Path (r) is 8 and Half of the Distance between the Centre Lines of the Wheel (a) is 4.

This implies that;

m = Mass of the Vechicle = 12

g = Acceleration due to Gravity = 9.8

v = Velocity of the Vehicle = 28

h = Height of Centre of Gravity of the Vehicle from Ground Level = 16

r = Radius of Circular Path = 8

a = Half of the Distance between the Centre Lines of the Wheel = 4

R_{B} = ^{mg} / _{2}[1 + ^{v²h}/_{gra}]

R_{B} = ^{12 x 9.8} / _{2}[1 + ^{28² x 16}/_{9.8 x 8 x 4}]

R_{B} = ^{117.60} / _{2}[1 + ^{784 x 16}/_{313.6}]

R_{B} = 58.80[1 + 40]

R_{B} = 58.80[41]

R_{B} = 58.80[41]

R_{B} = 2410.8

Therefore, the **reaction of the outer wheel of a vehicle moving along a level circular path** is **2410.8 N.**