How do I determine the force and torque requirements for a 12V DC car jack?

[Othman Safarini]

< Moderator Note -- thread moved from ME forum to Homework Help/Engineering >[/color]

Hi,

I Am an electrical engineering student and I want to make a 12 DC car jack for my senior project, but till now I couldn't find the informations I need to start the project.

How to calculate the force and torque needed to lift the car ?
And which one is better : hydraulic bottle jack or scissor car jack for the project ?

 

[Dr.D]

When you say, "12 DC" are you referring to 12 volts DC?

Have you taken statics? That is what you need to calculate the force required to lift the car. Without that knowledge, you will be ill equipped (even dangerous) to take on this task.

 

[Othman Safarini]

When you say, "12 DC" are you referring to 12 volts DC?

Have you taken statics? That is what you need to calculate the force required to lift the car. Without that knowledge, you will be ill equipped (even dangerous) to take on this task.

Yes , I'm referring to 12 volts DC.
The idea of the project is to build an electric car jack that powered by 12 volt electricity supplied directly from the car's cigarette lighter receptacle.

 

[Mark44]

One thing that might be a concern is that the cigarette lighter is in a circuit protected by a fuse. In my car, for example, the lighter circuit has a 7.5A fuse. If I were to hook up a motor drawing, say 10A, it would blow this fuse in my car.

 

[billy_joule]

What's the answer to Dr D's second question?

Have you used a bottle jack and scissor jack? Which do you think would be easier to power? Can you calculate available power from the lighter outlet?

 

[Othman Safarini]

One thing that might be a concern is that the cigarette lighter is in a circuit protected by a fuse. In my car, for example, the lighter circuit has a 7.5A fuse. If I were to hook up a motor drawing, say 10A, it would blow this fuse in my car.

We are aware of this problem , we will find the solution for this problem after we choose the proper DC motor for the project.
Now we are concentrate on choosing the car jack.

 

[Othman Safarini]

What's the answer to Dr D's second question?

Have you used a bottle jack and scissor jack? Which do you think would be easier to power? Can you calculate available power from the lighter outlet?

I used the scissor car jack but the bottle jack I didn't use it till now.
We think that the scissor jack is easier to power.
We will do the calculation of the output from the lighter outlet after choosing the type of car jack.

 

[Dr.D]

He can always keep the power requirements low enough if he is willing to wait long enough to raise the car.

 

[CWatters]

The force required to lift one side of a car is going to be about half the weight of the car. So for something like a Dodge Durango the force will be something like 15,000N (3000Kg * 9.81 / 2). It's not easy to convert that to torque because you don't know the friction in the jack.

The easiest way to work out the torque might be to actually lift a few cars using a torque wrench on the screw jack? Plot a graph of torque vs weight? Scale to the max likely weight? Add some contingency? Something like that.

 

[CWatters]

Lets say you are limited to 5A at 12V = 60W. The fastest you could raise a car like the Durango would be..

Power = force * velocity
velocity = power/force
= 60/15000
= 0.004 m/s

If you need to raise it say 0.25m it would take about 60 seconds. Which appears reasonable.

Might be possible using a cheap Mabuchi motor with suitably robust gearbox. I think the gearbox is likely to cost more than the motor.

Edit: I haven't allowed for the inefficiency of the motor/gearbox/scissor jack which could be considerable. If you assumed 50% ?? that would double the time but still ok.

 

[Othman Safarini]

Lets say you are limited to 5A at 12V = 60W. The fastest you could raise a car like the Durango would be..

Power = force * velocity
velocity = power/force
= 60/15000
= 0.004 m/s

If you need to raise it say 0.25m it would take about 60 seconds. Which appears reasonable.

Might be possible using a cheap Mabuchi motor with suitably robust gearbox. I think the gearbox is likely to cost more than the motor.

Edit: I haven't allowed for the inefficiency of the motor/gearbox/scissor jack which could be considerable. If you assumed 50% ?? that would double the time but still ok.

Thank you for your help.

If we want to use the hydraulic bottle jack, can we find the torque by find the force needed to lift the car ?

 

[CWatters]

If we want to use the hydraulic bottle jack, can we find the torque by find the force needed to lift the car ?

Torque isn't relevant to bottle jacks. At least not the ones I've used. Pressure is a more relevant concept. Yes you can calculate the pressure needed by finding the force needed to lift the car. Other information is also needed.

 

[Othman Safarini]

Torque isn't relevant to bottle jacks. At least not the ones I've used. Pressure is a more relevant concept. Yes you can calculate the pressure needed by finding the force needed to lift the car. Other information is also needed.

Yesterday we found a new type of car jacks which is a hydraulic bottle jack but with a screw instead of the bumb just like the scissor jack.
In this case how can we convert the pursuer and force to the torque needed from the motor ?

 

[Jobrag]

For the kind of power coming from a lighter socket my gut feeling is that it will take so long to lift the car that good old muscle power would be better. Having said that have you considered an air bag with an electric fan?

 

[CWatters]

Yesterday we found a new type of car jacks which is a hydraulic bottle jack but with a screw instead of the bumb just like the scissor jack.
In this case how can we convert the pursuer and force to the torque needed from the motor ?

Can you post a link? I've not seen that type.

This page might help..
http://technologyinterface.nmsu.edu/3_2/3_2f.html

 

[Othman Safarini]

Can you post a link? I've not seen that type.

This page might help..
http://technologyinterface.nmsu.edu/3_2/3_2f.html

Like this one

http://sell.lulusoso.com/selling-leads/797701/screw-jack-dome-jack-car-jack.html

 

[billy_joule]

Yesterday we found a new type of car jacks which is a hydraulic bottle jack but with a screw instead of the bumb just like the scissor jack.
In this case how can we convert the pursuer and force to the torque needed from the motor ?

Without knowing the efficiency and velocity ratio of the mechanism you can't.
I'd guess it's a small rotary positive displacement pump. Much less likely is a power screw (like a scissor jack).
You would have to estimate losses to get a ballpark figure, You will likely get a much better figure doing it experimentally like CWatters link does. Have you talked to your supervisors about this? A senior project in something you know so little about is not a good idea. This is much more a ME project than EE. The only thing electrical about this is motor selection and ME students should know enough about that to do it well. In fact, this is the exact type of question ME's will do a lot of in design papers.

 

[CWatters]

What billy said.

The question you are asking can only be answered by the manufacturer of the jack. It's like asking "how much force can a lever generate" when you don't know the length of the lever.