Energy required to lift 1 pound 3 inches at 70 MPH

[motolectric]

Hi,

I have a few energy questions and I'm hoping they are welcomed at this forum.

I'm dealing with the idea of reducing "unsprung weight" on motorcycles.

This is the weight of items below the suspension, i.e. the wheel, tire, brake discs etc.

My first foray into this was back in the Usenet days when I asked on a Usenet forum why reducing 1 pound from my tire (by going to a smaller cross section tire - and 25 inches diameter to 24) made such a huge difference to how my motorcycle handled. The guy who answered said that since the weight difference was located 12 inches from the axle that it was the equivalent of a 64 pound reduction in weight (I had supplied the data in my question that the only difference between the tires, was 1 pound in weight which seemed [based on the appearance and measurements] to be all at the circumference of the tire).

His figure coincided with my seat of the pants conclusion, i.e. it was very large.

I state the above in case anyone here can chime in that it is off or add anything to it.

My question now is;

I have a wheel/tire/fender/axle that is going along at 70 MPH and it hits a bump that is 3 inches high.

I have another wheel/tire/fender/axle that weighs 1 pound less and it is going at 70 MPH and it hits the same bump.

How much less energy is needed to lift the 2nd combination vs. the 1st?

On motorcycles reducing unsprung weight makes very large differences in the handling but the changes are on the order of 8 ounces, 16 ounces etc.

I'm curious as to how much energy savings there are and why.

Also is there a curve to the energy required to lift the 2 weights (I would think there is based on the speed).

If the calculations need real world base figures I would think 40 pounds and 41 pounds would be close to what is on a motorcycle.

But would it be different if the real world weights were 1 pound and 2 pounds?

Thanks for any tips/advice/feedback.

M./

 

[Drakkith]

How exactly did your motorcycle handle differently?

 

[motolectric]

The difference was an increase in the ease of leaning the bike over in curves. In rider parlance it became "more flickable". Was obviously due to a reduction in the gyroscopic forces but the feel of the difference did not equate with the small reduction in weight or overall diameter of the tire. The 64 pound figure did feel closer to what I felt on the bike, i.e. a large difference.

Thanks for any tips/advice/

M./

 

[skeleton]

-------------------------------------------------------------------------

A rough road (wallows, dips, etc) will impart a vertical acceleration on the front wheel of the bike. If the acceleration is lower than what the spring can absorb, then the shock will take up all of the impact. (GOOD) If the acceleration is greater, then the bike will lift up (impact) by the amount of the excess acceleration.

So, what is the acceleration that the road applies to the front wheel?

This is a geometry question. Consider the bump is defined as an equilateral triangle of height 3 inches. Draw the triangle at the base of a line. Beside it draw a circle corresponding to the diameter of the wheel. As the bike rolls forward, the wheel has to ride up the triangle, until the wheel is directly over the triangle. The horizontal speed is known, 70 mph. The vertical height is known, d=3 inches.

You need to measure the outer diameter of your tire.

Now, apply some geometry with the sketch above, to determine the vertical speed of the wheel, at the point when it has risen to the top of the triangle.

Now, you can estimate the vertical acceleration of the wheel:

a_vert = (v_vert)^2/2/d

-------------------------------------------------------------------------

Reference
http://en.wikipedia.org/wiki/Harmonic_oscillator

Estimate your unsprung mass. Convert weight (40 lbs) into mass.

Use a stop watch and measure the time it takes the front suspension to be compressed and then rebound. It is immaterial how much displacement the shocks are compressed. This full time is one cycle, and gives the spring's period of vibration, T.

Given
T = 2*pi*sqrt(m/k)
T = period of oscillation of the spring (compression and rebound cycle)

Then
k = m*(T/2/pi)^2

From the above calculation, estimate the spring constant, k.

-------------------------------------------------------------------------

Reference
http://en.wikipedia.org/wiki/Spring_(device)

Given #1
F = m*a
m = mass of unsprung wheel, 40 lbs
a = acceleration of wheel, as it moves vertically up.

Given #2
F = -k*x
k = spring constant of the front forks
x = vertical compression of the front forks, 3 inch.

Then
a_shock = k*x/m
a_shock = maximum acceleration that shock can absorb.

The above equation gives the maximum vertical acceleration that the spring can compress at. If a faster acceleration is applied to it from the rough road surface, then the WHOLE BIKE will jump up at an acceleration rate being the difference of what is applied and what the shock can absorb.

Thus
a_bike = a_road - a_shock

-------------------------------------------------------------------------

Conclusion

Obviously, a rider wants the shock to absorb ALL of the vertical wheel acceleration. That way, the bike does not lift up. If it were to lift up, then it takes a relatively long time for the bike to fall back down - during that time, the wheel is lost some degree of contact with the road. This is where traction is lost.

Ideally, we want:
a_bike = 0

In this case,
a_shock=a_road
a_shock=k*x/m

If unsprung weight went from 41 lbs to 40 lbs, then:
delta(a_shock) = (1/40-1/41)/(1/41)
=2.5%

So, a 1 lb reduction in wheel weight will allow 2.5% increase in applied vertical acceleration (bump).