# Maximum acceleration for a vehicle [closed]

I'm in engineering school and we have a project: we have to build a amphibioues vehicle; I'm looking for a formula.

Our vehicle has to go as far as possible with its unique source of energy, a spring. But if this spring is very powerful then our vehicle might skid.

I searched and found a formula that related the best velocity to go as further as possible :

a_max = μg

With:

• $R$ the radius of a wheel
• $\mu$ the coefficient of friction
• $m$ the mass
• $I$ the moment of inertia

The problem is we are looking for a velocity (without t) to calculate the ratio to transmit to the gear system.

We found the equation for the acceleration but not the velocity, that what we are looking for.

• I don't understand. You've given a formula for acceleration, but say that you are looking for a velocity. What velocity are you looking for? I don't think velocity has any role in predicting whether or not the vehicle will skid. – garyp Mar 19 '15 at 21:09
• See related post physics.stackexchange.com/a/15620/392 to find distance traveled when acceleration is defined as a function of distance (spring deflection). – ja72 Mar 20 '15 at 20:39
• Peak acceleration is most likely to happen at zero velocity, at launch. – ja72 Mar 20 '15 at 20:48
• Maybe you should put your source for the formula so we have some background. – Reid Erdwien Mar 20 '15 at 21:49

I don't think that equation is right. $F_{max} = \mu mg$, so $a_{max} = \mu g$.

Where are you getting velocity from? The spring doesn't move at a constant velocity, does it? You need to use the spring's maximum torque and work out how to weaken it so the final acceleration is sufficiently low.

Why are you trying to make it as fast as possible? If you're going for distance, you have to minimize losses due to friction. Slower would generally be better, although at some point the friction from the gearbox will cause problems.

• Yes your right we are trying to go as further as possible and not the fastest. We can say that the spring move at linear velocity – user1892467 Mar 20 '15 at 19:42
• Are you sure the spring is going to move at the same rate regardless of how much force it has to exert? That seems unlikely. – DanielLC Mar 20 '15 at 19:59
• Yes your right, but for the moment I don't have the mathematical description of the force of the spring. Do you think its essential in finding the relation between velocity and acceleration ? – user1892467 Mar 20 '15 at 20:04
• I think what you should do is try it for different gear ratios and see what works. You're using a lot of models that are only approximately accurate, so calculating it mathematically will only be so helpful. For example, I've seen people fix a robot that was slipping down a hill by adding weights. If you plug that into your equations, you'll find it shouldn't matter. If you insist on working it out by hand, you'll need a better model of the spring. I think that the bottleneck will be at the very beginning, so see how much force it exerts when fully stretched, and use that. – DanielLC Mar 20 '15 at 20:15

If the vehicle has 4 wheels, and only 2 have traction then the formula is

$$\begin{cases} a_{max} = \gamma \mu g & \mbox{RWD}\\ a_{max} = (1-\gamma) \mu g & \mbox{FWD} \end{cases}$$

where $\mu$ is the traction coefficient, $g$ is gravity and $\gamma$ is the %weight on the back wheels. So if the weight distribution is 60%/40% front to back, then $\gamma=0.4$

To get a level up in detail, if the height of the center of mass to the wheel axle is $h$ and the wheelbase is $\ell$ then the maximum acceleration is

$$\begin{cases} a_{max} =\frac{\ell}{\ell-\mu h} \gamma \mu g & \mbox{RWD}\\ a_{max} = \frac{\ell}{\ell+\mu h} (1-\gamma) \mu g & \mbox{FWD} \end{cases}$$

If you have All Wheel Drive (AWD) then $a_{max} = \mu g$.

I got to this with a crude free body diagram and the equations of motion. • Thanks a lot, for information the vehicle is in traction. And yes we are trying to go the further as possible and not the fastest. But the problem was how to get a velocity from the acceleration (without a time variable) ? – user1892467 Mar 20 '15 at 19:30
• You want the power produced by the spring $P=F(x) \dot{x}$ to equal to the power of acceleration (change in kinetic energy) $P=(m v) a$. Find the relationship between spring flex rate $\dot{x}$ and speed $v$ and equate the two powers. $$P = \frac{{\rm d} }{{\rm d} t} \frac{1}{2} m v^2 = m v \frac{{\rm d}v}{{\rm d} t} = m v a$$ – ja72 Mar 20 '15 at 20:11
• If your motor (spring) is making $P=100$ Watts and the vehicle weight is $m=10$ kg then at $v=1$ m/s it will accelerate by $100 = (10)*(1)*a$ or $a=10$ m/s^2 – ja72 Mar 20 '15 at 20:14
• Yes ok but how can I exploit it to find the velocity ? – user1892467 Mar 20 '15 at 20:24
• @user1892467 the op question asks about max. accelerations. The comments is no place to answer a secondary question. Please award this question and ask a new one on how to use max traction to design the transmission mechanism for optimal distance. This is not that simple as more information is needed than supplied in this posting. – ja72 Mar 20 '15 at 20:28

You cant

avoid skidding by an equation.

Would You glue a paper with that formula to Your car?

You avoid skidding by using the appropriate tires for the floor.

Look here for a solution for a car going as far as possible:

en.wikipedia.org/wiki/Mousetrap_car