Chescadence
Guest
Ok, so we've all heard the good old' "1lb on the wheels/tires counts of 2lb on the frame"..
This can be quite easily calculated, total Kinetic energy of a wheel is the linear translational energy + the rotational energy, so:
Ktrans= 1/2xmv^2 and Krot= 1/2xIomega^2 - If we for arguments sake say that a wheel is a cylindrical shell with all it's weight at the rim (Not exactly true but..); we could say the Inertia is I=MR^2. We know Omega (Angular Velocity)= V^2 / R^2. We can then adjust this formula to work out the overall kinetic energy for a wheel and substitute in out I and Omega values:
Ktrans= 1/2xmv^2 , Krot= 1/2xMR^2xV^2/R^2 = MV^2/2 (After simplification, R^2's cancel each other out and are not important in this sum and 1/2xMV^2 is the same as MV^2/2)
Lets assume that our bike wheel has a mass of 1.5kg and is moving at 20kph= (20x10^3/360) = 5.555555ms^-1
KEtrans = 1/2x1.5x(5.5555555)^2 = 23.15 joules , KErot= (1.5X5.5555555^2)/2 = 23.15 joules
KEoverall = 46.30 Joules. From this we can deduce that the KE for the wheels is twice that trans/linear Energy if the weight was on the frame due to the lack of rotational KE - Not exaclty x2 due to the spokes etc. So essentially, if we put 1.5kg on a wheel you'd require nearly twice the energy to power it than if it was on the frame.
So lighter wheels would make acceleration easier for sure, but at a constant speed would the wheels KE still "seem" x2?
I'm studying Meds so i'm by no means a physicist so i'd be interested in anyone with greater expertise than me.. Hope people got the gist anyway.
This can be quite easily calculated, total Kinetic energy of a wheel is the linear translational energy + the rotational energy, so:
Ktrans= 1/2xmv^2 and Krot= 1/2xIomega^2 - If we for arguments sake say that a wheel is a cylindrical shell with all it's weight at the rim (Not exactly true but..); we could say the Inertia is I=MR^2. We know Omega (Angular Velocity)= V^2 / R^2. We can then adjust this formula to work out the overall kinetic energy for a wheel and substitute in out I and Omega values:
Ktrans= 1/2xmv^2 , Krot= 1/2xMR^2xV^2/R^2 = MV^2/2 (After simplification, R^2's cancel each other out and are not important in this sum and 1/2xMV^2 is the same as MV^2/2)
Lets assume that our bike wheel has a mass of 1.5kg and is moving at 20kph= (20x10^3/360) = 5.555555ms^-1
KEtrans = 1/2x1.5x(5.5555555)^2 = 23.15 joules , KErot= (1.5X5.5555555^2)/2 = 23.15 joules
KEoverall = 46.30 Joules. From this we can deduce that the KE for the wheels is twice that trans/linear Energy if the weight was on the frame due to the lack of rotational KE - Not exaclty x2 due to the spokes etc. So essentially, if we put 1.5kg on a wheel you'd require nearly twice the energy to power it than if it was on the frame.
So lighter wheels would make acceleration easier for sure, but at a constant speed would the wheels KE still "seem" x2?
I'm studying Meds so i'm by no means a physicist so i'd be interested in anyone with greater expertise than me.. Hope people got the gist anyway.