Maximizing Speed with Front and Back Wheel Ratio in a Spring Powered Dragster

  • Thread starter Thread starter klkc
  • Start date Start date
  • Tags Tags
    Spring
klkc
Messages
4
Reaction score
0
1. i am in my first year of my engineering course, now doing a project that involves designing, building and testing a spring powered dragster. i am charged with the design of the wheels. so far i know that inertia is proportional to the square of the radius (k and m almost constant) and the spring has stiffness of 2.01 N/mm.
what equations are there that relate inertia to angular acceleration, how does the ratio front wheel ratio: back wheel ratios (apart from shifting the weight distibution), and what would k be for a wheel (im thinking 6 spokes equally spaced)?

thanks
 
Physics news on Phys.org
> what equations are there that relate inertia to angular acceleration

You mean moment of inertia? \tau=I\alpha (that is, torque, or moment, is equal to the moment of inertia multiplied by the angular acceleration)

> how does the ratio front wheel ratio: back wheel ratios (apart from shifting the weight distibution)

That isn't a complete question.

> and what would k be for a wheel (im thinking 6 spokes equally spaced)?

k, as in the radius of gyration? k=\sqrt{\frac{I}{m}}
You can either perform the integrals yourself or look up the moments of inertia for the objects that make up a wheel (and just sum them together).
 
what i meant was does the choice of the size of the front wheel affect the choice of the size of the back wheels if one wants to maximise speed and i also would like to know how they are related. also kinda confused. i think in order for the wheel not to slip the moments around the centra of it must balance but then there would be no angular acceleration as there is no unbalanced moment.
 
The wheels slipping depends on the horizontal force exerted on the ground and the maximum static friction force the ground can supply.

>does the choice of the size of the front wheel affect the choice of the size of the back wheels if one wants to maximise speed

I'm not sure. I suppose the rear wheels on dragsters are larger because that would increase the surface area in contact with the ground (which would increase the maximum frictional force), although this only makes sense if the rear wheels are the only wheels with a torque applied to them. A larger wheel would also probably mean a larger moment of inertia, however, which would mean a slower start but would also mean it would have more angular momentum.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top