Unique GoKart need some help (spring assisted)

[nacanitihs]

I am a mechanical engineering major and for our senior project we are to design a spring assisted/powered vehicle. we are going to use a go kart as a base vehicle. i would like some assistance from some you guys.

i am trying to figure out how much torque from the spring would be required to move the go kart. and how long the power would last. so i can select the proper size.

anyway i have made a few calculations

Drive gear 1.5in
Rear axle drive gear 8in
Tire size 11.5in
weight of kart and person 350lbs

T=Fr
T=350lbs*(5.75in*0.75in)/4in
T=377.34inlb

now i looked at gokartguru.com and the guy was calculating motor size to go up a 10 degree hill. he multiplied a similar torque by sin(10) now that would give the verticle side of the triangle and that brings the torque to T=65.5 inlb
and he based the size of the motor off of that.

doesnt that seem a little too small? am i missing something?


Ill be updating this thread as the progress of the kart continues.

 

[blackparis]

your equation is messed up.. if you ask me. (its right but not the one you need)

you would need a target rate (mph or kph), and your gear sizes are going to have to show circumferences, rather than radii....

I'll try to come up with something for you and post back.

Also whats going to be your method of winding this spring? Is it just "pull the vehicle backward?"

 

[t0x1k]

Weld a small spring to the top of the piston and use that engine on the go kart. Call it the piston return spring helper.
http://kalecoauto.com/index.php?main_page=product_info&cPath=3&products_id=27



JOKE!

 

[blackparis]

Ok.. I spent some time on this.

Lets say you had a 1inch solid axle in the back.. Forget about changing the gear ratios at the moment and you wraped a rope around it two revolutions(6.28inches, pi*2) you have 11.5 inch tires on this axle.

if you pull the cord, your gearing advantage is 1 : 11.3 so you cart will move 71.02 inches under power. plus whatever it rolls from momentum. which is determined by the force use to pull the rope - minus any resistance to rolling (grade, rough surface, aerodynamic losses, etc)


Obviously, the rolling resistance of 11.5 inch tires on a smooth flat surface, isn't much. thus your only need is to calculate howmuch force you need for an optimum propulsion (not doing a wheelie and causing a backflip(or doing a burnout), while building the most momentum)

.... I dont want to do your project for you...

LOL.. just want to give you some ideas.. I'll have more coming in an hour or so. gotta run to the store. BRB

 

[blackparis]

Ok.. So now lets figure out the force we will need to make this roll... lets say a 350lb cart with 11.5 inch tires has a rolling resistance of 30lbs on a flat surface. (exagerated) But at the pull string we have a gearing disadvantage. 11.3 : 1 So NOW we need ~330lbs to over come being stationary... Thats a pretty nasty tug.. LOL.

OK.. so if we replaced the tug, with a compressed (6.28inches) 330lb spring, we might actually get going pretty well... 330lbs over 6inches is 1980 total lbs.

The next step would be figuring out your acceleration rate. and the speed you will be traveling when momentum takes over. I don't feel like doing anymore for you, I'm not the one getting a degree.. LOL

Keys points are:
this was just an example.
Wind it more than twice.. 6inches isn't very much.. LOL. But then gotta find a spring that can pull or push a farther distance.
Bigger harder tires will decrease rolling resistance.
If spring rate is too high.. Burnout or Wheelie.. LOL.. Sound fun!
Higher gear ratios will increase torque.. lower speed.. more probable wheelie...
Wheelie bar with low resistance rollies, might be nescasary

 

[GoKartGuru]

10 degree Hill Factor: What Is That?!

I can see where people would look at that first torque equation and go "huh?!"

The torque value that moves a go kart is dependant on the rolling friction. In the calculations that are presented on the gokartguru.com page, the friction is assumed zero, so the force that is being calculated is actually due to an accelleration.

The 10 degree hill is actually what will stall most go karts, because they are designed to just go on flat surfaces. The 10 degree calculation adds in a weight that the engine will see. Which in this case is the sin of 10 degrees or approximately 17% of 350 or 60 lbs

The torque that the rear axel would see on a hill is based on this number of 60 lbs. In otherwords, the rear tire has a diameter of 11.5 inches or approximately 6 inch radius. The torque that the rear axel would see is 60lbs x 6 inches or 364 in-lbs. The torque at the drive gear is related to the tension of the belt or chain, which obviously is related to the sprocket diameter.

To calculate the tension in the chain figure out the force sprocket face (T=FR, solve for F.....F = T/R) Use this tension to calculate the torque at the drive sprocket.

So Force Chain = 365 in-lbs/ Rmain drive sprocket (4 inches) = 92 lbs

Use this force to calculate torque at the drive sprocket (T = Force Chain x Radius Sprocket)

Torque engine = 92 lbs x .75 inches = 69 in-lbs.

Your original assessment was probably correct. The torque being shown here is the amount of torque that would balance the go kart or stall the engine when climbing a 10 degree hill.

You obviously need more torque if you don't want the engine to stall. The purpose for this calculation is to assess what is the minimum amount that would be acceptable. This torque of 69 in-lbs would move the go kart okay on a flat surface, but would cause the go kart to smoke the clutch and stall on a hill.

A better calculation is to shoot for an accelleration of about 1/4 g (or 8 ft/s^2) and then compare these numbers with the 10 degree hill torque value. If the 10 degree value is higher, then you need to up the horsepower on your engine.

Hope this helps.....