The Isle of Man TT is a truly unique motorcycle competition

The Isle of Man TT is a truly unique motorcycle competition. You plan on attending the
event with a CBR1000RR, data listed in the following table.

Initial preparation for the competition requires you to answer the following questions.

a. Sag is motorcycle technical jargon that defines the static deflection of the front and rearsuspension. At the front it is measured as fork travel from full extension and at the rear it ismeasured as displacement of the rear wheel centerline from full droop. Calculate the amount of preload in inches required so the front and rear sag equal 1”.

b. Including the deflection of the tires, what is the static deflection in inches of the front and rearsuspension of 1a.

c. Ignoring the influence of preload, what is the static deflection in inches of the front and rearsuspension. Compare to 1b and comment.

d. What front and rear springs are required for 1c to equal the deflection of 1b.

e. If the motorcycle is travelling at a constant velocity of 90mph, what is the radius of a crest sothat the static deflection of the motorcycle equals zero.

f. If the motorcycle traverses a dip of the same radius as 1e, what is the total compression for thefront and rear suspension of 1a?

Curb Weight 615 lbs

Wheelbase 55.5 in

Weight Distribution 48.0%

Unloaded Center of Gravity Height 32.145”

Caster Angle 23.0°

Front Spring Rate 60 lb/in

Front Stroke from FullExtension 5.25”

Front Un-sprung Weight 23.5 lbs

Front Tire Spring Rate 860 lb/in

Front Tire Diameter 23.6 in

Front Tire Section Width 4.75 in

Front Tire Inertia, Iwf 30.357 slug-in2

Rear Spring Rate 675 lb/in

Rear Motion Ratio 1.75

Rear Stroke of Axle Centerlinefrom Full Extension 6.5 in

Rear Un-sprung Weight 32.5 lbs

Rear Tire Spring Rate 740 lb/in

Rear Tire Diameter 24.8 in

Rear Tire Section Width 7.09

Rear Tire Inertia, Iwr 114.047-in2

Gravity Constant 32.2 ft/s2

1 lbf 1 slug-ft/s2

Solution

GIVEN:- Curb Weight (W)615 lbs, Wheelbase (wb)55.5 in, Weight Distribution 48.0%, Unloaded Center of Gravity Height (h)32.145”, Caster Angle 23.0°, Front Spring Rate (Fsr)60 lb/in, Front Stroke from FullExtension (FS)5.25”, Front Un-sprung Weight (Fw)23.5 lbs, Front Tire Spring Rate (Ftr)860 lb/in, Front Tire Diameter (Fd)23.6 in, Front Tire Section Width (Ftw)4.75 in, Front Tire Inertia, Iwf 30.357 slug-in2, Rear Spring Rate 675 lb/in (Rsr), Rear Motion Ratio 1.75, Rear Stroke of Axle Centerlinefrom Full Extension (RS)6.5 in, Rear Un-sprung Weight (Rw)32.5 lbs, Rear Tire Spring Rate(Rtw) 740 lb/in, Rear Tire Diameter (Rd)24.8 in, Rear Tire Section Width(Rtw) 7.09, Rear Tire Inertia, Iwr 114.047-in2, Gravity Constant 32.2 ft/s2

SOLUTION:-

(a) Considering weight distribution of 48%, weight acting on front tire = 615 x 0.48 = 295.2 lb and weight acting on rear tire = 615 - 295.2 = 319.8 lb

For front suspension ---> 295.2 = 60 x deflection, So deflection = 295.2 / 60 = 4.92 in, but since static deflection required fsd = 1 in, so preload required in inches Pfront = 4.92 - 1 = 3.92 inches (ANSWER)

For rear suspension ---> 319.8 = 675 x deflection, So deflection = 319.8 / 675 = 0.47 in, but since static deflection required fsd = 1 in, so preload required in inches Pfront = 0.47 - 1 = -0.53 inches (ANSWER)

(b) Considering front tire spring rate, deflection in front tyre = 295.2 / Ftr = 295.2 / 860 = 0.34 in

Total static deflection in front suspension = 0.34 + 1 = 1.34 in (ANSWER)

Now, considering rear tire spring rate, deflection in rear tyre = 319.8 / Rtr = 319.8 / 740 = 0.43 in

Total static deflection in rear suspension = 0.43 + 1 = 1.43 in (ANSWER)

(c) Ignoring preload in 1(a), the total static deflections are as follows-

Front suspension 0.34 + 4.92 = 5.26 in (ANSWER) (Comment - Deflection increased in absence of preload)

Rear suspension 0.43 + 0.47 = 0.9 in (ANSWER) (Comment - Deflection decreased in absence of preload)

(d) With respect to obtaining static deflections mentioned in 1(b), the springs required are as follows -

Front suspension spring with Spring rate = 295.2 / 1.34 = 220.30 lb / in (ANSWER)

Rear suspension spring with Spring rate = 319.8 / 1.43 = 223.64 lb / in (ANSWER)