0
\$\begingroup\$

I am trying to make an electric go-kart. I am new to the forum, so I may be missing some information but any help is greatly appreciated. After doing some math and discussing with friends I found the following with the given information. I have 4, 24V 14A motors, all connected to a read drive axle. I have 2 pairs of motors on each side of the drive axle. The motors have an 11 tooth sprocket that is chained to a 68 tooth sprocket that is connected to the drive axle. I currently have 2, 12V 7Amp 20 hour lead-acid batteries, which provide my 24V. Each pair of motors is connected in parallel to ensure they both get the 24V. However, the motors do not spin, even when there is no load (The wheels aren't contacting the ground). I believe that this may be because the motors do not have enough amps, only receiving 7 Amps of the 14 Amps from the batteries, because there are 2 motors in each pair. My main goal is to get the kart moving. I weigh about 130lbs, and the kart is likely around 30lbs. Would getting a battery/batteries with move amps (28A) allow the motors to run properly? Any advice is really appreciated.

All the information I know

Each motor is a DC motor that needs 14A and 24V The motor has a sprocket of 11 teeth The Watts of each motor is 250 The motors are rated to have an RPM of 2750 or around 45.83Hz

The drive axle has a sprocket of 68 teeth. The 68 toothe sprocket is 5.5" The wheel diameter is 10"

If my math is correct, given the ratio of the motors to drive shaft. Each motor produces 5.45Nm of torque, and after the sprockets, the torque is 33.69Nm. I don't know what impact the 4 motors would have on the torque, nor do I know what the torque is for each side. I really just want this kart to move.

Would removing motors allow my vehicle to move? I think that my power supply does not provide enough amperage to my motors. Would having 1 or 2 motors be better in this situation? enter image description here

Data sheet for my batteries

\$\endgroup\$
9
  • \$\begingroup\$ 50 W would move the cart at low acceleration. Your question is vague. (1) Do the motors move when bench tested? (2) What is the motor rated speed and what is the gearbox output speed then? (3) What is the torque rating? (4) What is the current draw on the bench and on the cart? Edit the question and add the information in. \$\endgroup\$ – Transistor Sep 22 '20 at 20:32
  • \$\begingroup\$ Compare the rated speed (or unloaded speed) of the motor with the speed you want on the axle. I bet you find the ratio is much more than 68:11, in which case you need to gear down a lot more, to avoid stalling the motors. \$\endgroup\$ – user_1818839 Sep 22 '20 at 21:08
  • \$\begingroup\$ In additions to the questions already -- do the rear wheels move when they're off the ground? Have you measured the current to the motors when they're running? Do you have any sense of whether the motors are trying to run? Can you share the motor data sheet (preferred), or a link to it? As mentioned, edit your question to include this information. \$\endgroup\$ – TimWescott Sep 22 '20 at 21:13
  • \$\begingroup\$ Well my 1:10 RC Car has a 350W Motor... 250W in a go-kart doesn't sound like fun to me \$\endgroup\$ – schnedan Sep 22 '20 at 21:17
  • 1
    \$\begingroup\$ It is really hard to debut this remotely. One thing you can do, though, is jack up the whole go kart, and try to get the wheels moving by giving them a push with your hand. Once they get started, maybe the motor will be able to spin them. Or maybe the brake is on. Or maybe the gear ratio is no good. Maybe you need a smaller gear on the motor shaft or a larger one on the drive shaft. You should probably do some calculations based on torque, power, etc. \$\endgroup\$ – mkeith Sep 23 '20 at 1:23
1
\$\begingroup\$

Properly geared and powered a 250 Watt motor will move a gokart at modest speed. Probably 5-10 kph and maybe more. More Watts does better. At slow speed power increases about proportionately with speed. As speed picks up and wind resistance matters power needed increases with cube of speed !

Wheel diameter influences driving torque.
The power source must be able to supply 10A at 24V for about 250 Watts of power. More for more.
You need to specify wheel diameter and power source.

Watts ~= kg.m torque x RPM
For 250 Watts torque in kg.m = Watts/RPM
Wheel RPM is Motor RPM x gear ratio.
= 2750 x 11/68 ~= 440 RPM.
So torque is ~= Watts/RPM = 250/2750 =~ 0.1 kg.m
If wheel diameter is say 12 inches = 0.3m
The force at road surface = Torque/radius
= 0.1/0.15 =~ 0.66 kg force.

That's "rather small".
It may be that your reduction drive requires more force than that to overcome static stiction. If you prop it up and power it and hand turn the wheels, does it spin?

You need substantially more down gearing.
Your power supply capacity needs to be adequate.


Derivation of the above "approximate power" formula:

The formula Power = Torque in kg.metres x RPM is an approximation that is about 2.5% low.

It works because various things just happen to cancel.
Power = Force x distance
= (kg x g) x 2.Pi.R x RPS = kg x R x (g x 2.Pi x RPS) versus
. kg x R x (rpm)
So we need g x 2 x Pi x RPs to equal RPM
= 9.8 x 2 x 3.14 = 61.544 RPS = 1.0257 RPM

so Kgm x RPM is ~= 2.5% low.

Despite the slight' inaccuracy this is an immensely useful formula for in-the-field use.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Indeed this seems an issue of unreasonable mechanical design. Such a low powered motor is going to need an additional reduction stage, and common bicycle style roller chain is probably not suitable for the speeds of the first stage. A belt might work there, but I suspect that things comparable to the askers goal such as "power wheels" toys probably use a gearbox at least for the first stage. Overall this is really a mechanical not electrical engineering problem; the question would be if it's complete enough to be eligible for migration or if it simply needs to be closed. \$\endgroup\$ – Chris Stratton Sep 23 '20 at 19:15
  • \$\begingroup\$ Is that really the right calculation? I have always used power (W) = torque (Nm) x speed (rad/sec). I can't see that RPM and Kg would cancel to give the same result. \$\endgroup\$ – mkeith Sep 23 '20 at 22:43
  • 1
    \$\begingroup\$ @mkeith Yes - it's correct to about -2.5% but the reasons are 'naughty'. It works because various things just happen to cancel. Power = Force x distance = (kg x g) x 2.Pi.R x RPS. || Formula is kg x R x rpm . So we need g x 2 x Pi x RPs to equal RPM = 9.8 x 2 x 3.14 = 61.544 RPS = 1.0257 RPM so Kgm x RPM is~= 2.5% low. \$\endgroup\$ – Russell McMahon Sep 24 '20 at 10:42
0
\$\begingroup\$

If your motors are 24v, and you are placing the pairs of motors in series, doesn't your supply need to be 48v? Your existing circuit provides approx. 12v to a paralleled pair of 24v motors.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.