Here are some sums I've done (in proper units!), based on the numbers in Dieselmans 1st post in this thread.
The text after a % sign is just a comment.
>> m_tyre=10;
>> m_wheel=9.3;
>> Rim_Radius=(16*25.3e-3/2);
>> Rolling_Radius=Rim_Radius+(225e-3*55/100);
>>
>> % Modelling the wheel as a flat disc plus
>> % a concentrated mass at the wheel rim
>> % Assume a proportion of the wheel mass is in the disc (P_disc_mass)
>>
>> P_disc_mass=0.5; % Assuming a 50 / 50 split
>>
>> I_disc=0.5*m_wheel*P_disc_mass*(Rim_Radius^2);
>> I_rim=m_wheel*(1-P_disc_mass)*(Rim_Radius^2);
>>
>> % Assume tyre mass is all concentrated at the rolling radius
>> % this should over-estimate the inertia
>>
>> I_tyre=m_wheel*(Rolling_Radius^2);
>> I_TOTAL=I_disc+I_rim+I_tyre
I_TOTAL =
1.2750
>>% This is the rotational inertia in kg m^2
>>
>> % As the total kinetic energy of the wheel is
>> % 0.5 * m * v^2 + 0.5 * I * (omega^2)
>> % and v = omega x r
>>
>> % kinetic energy = (v^2)/2 * (m + I/(r^2))
>> % i.e., the effective mass is m + I / (r^2)
>>
>> Eff_mass=m_tyre+m_wheel+(I_TOTAL/(Rolling_Radius^2))
Eff_mass =
31.2862
>> % This is in kg
>>
>> Effective_Mass_Ratio=Eff_mass/(m_tyre+m_wheel)
Effective_Mass_Ratio =
1.6210
This means that the effect of the rotational inertia of the wheel is to make it seem about 60% heavier. Tweaking the proportion of mass at the rim of the wheel or on the face, via P_disc_mass will change the results a bit, but will not get anywhere near to 4x.
The big error in the spreadsheet is that it does assume that torque at the road wheel and torque at the engine are the same - where of course, there must be the overall gear ratio of the gearbox and final drive to take into account.