# Help! Physics Question...

#### TCG

##### New Member
Greetings...We have a discussion here and need some educated advice...

If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on 1 hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...and how much less mass have you moved over that same hour.

It's been slow at the office lately...

Thank you in advance for your time and consideration of our request.

Best regards from Japan...TCG

#### MNJRC_Berko

##### New Member
Ok, I cant solve this for one reason:

We need to know exactly how much energy it would take to complete one revolution of the crankarm WITH the extra 2 grams.

#### frey

##### New Member
Errm one physicist here, and sad though it is I couldn't resist replying In theory I don't think you'd save any energy at all, that is if you disregard friction and assume that you're not accelerating or going uphill (I am a physicist remember, we don't do real world!). You have to remember the no matter how heavy the pedal or shoe there's an equal weight on the other end of the crank, so it won't become any easier or harder to spin. Basically on the flat the only influence the weight of the bike has is on the friction between the moving components and between the tyres and the ground.

So if we assume that bike and rider weigh about 80kg, we can use the friction equation F=mu R, where R is the force resultant from the gravitational downforce on bike and rider, F is the frictional force, and mu is the combined friction coefficient, which we assume to be constant.
If we decrease the mass of the bike and rider by 2 grams, then we can say that F2/F1 = (79.998)/(80) = 0.999975 where F1 is the frictional force before massacering your shoes and F2 is the frictional force afterwards.
Since energy is E=Fd where F is the force being overcome (friction in our case) and d is the distance travelled, we find that Power is given by P=Fv where v is the velocity.
Since F2/F1 = 0.999975 we find that
P1/P2 = 0.999975 provided the same speed is assumed.
Put as a percentage you would save 0.0025% when it comes down to your power output.

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