Required voltage for torque specs

[elbeasto]

Hey all, I am trying to design a simple motorized grinder but I am running into a few problems. Just to give a background on myself, I am a computer major with no engineering education.

I need a motor that will have about 1-2 Nm of torque but be small and compact. I found the motor that would be perfect for my needs however one small problem. Those values were based on 48volts. This leads me to my question.

What is the limitation of a 9v battery in terms on torque? I am sure it has a lot to do with the design of the motor but if a 9v battery cannot produce that type of torque, what other options might I have?

Also, as a side note of the project, it is a simple grinder that will have run times of 15-20 seconds at a time.

Thanks

 

[Angello90]

I'm not too sure, but if you think about it;

\tau = d\vec{L}/dt and
\vec{L} = I\vec{\omega}

Now provided that on 9 volt battery revs are not going to be as high as on 48v power supply, than your \vec{\omega} will be smaller, thus the \tau will be smaller.

Correct me if that sounds wrong.

 

[elbeasto]

Thats actually exactly what I was scared of. I had a feeling I was going to get a response like that but unfortunately for me I do not even know what those symbols represent. My educational background is in networking and server management so that is quite literally greek to me.

However, I have no problem learning it if you wouldn't mind breaking down what each symbol is.

 

[Angello90]

So \tau is your torque,
\vec{L} is the angular momentum which is donated by I\vec{\omega}. Going further I is a moment of inertia and \vec{\omega} is angular velocity.

Since any kind of momentum is somehow connected to the speed, the value for \vec{\omega} is decreased, value for I is decreased, thus \vec{L} is decreased.

d\vec{L}/dt is the change of angular momentum with respect to time.

Let's say that at 48v the motor in a space of 1 second you have produces 2 N m s / s. Seconds cancels out and you're left with your torque 2Nm.

In the case of 9v, the motor won't be as fast as it was on 48v. There won't be enough current to drive it at the same speed as it did on 48v.Therefor instead of 2 N m s / s in 1 second you get a fraction of what you would produce on 48v, let's say 3/8 N m s / s in 1 second. And that's would only hold provided that decrease in power would be linear (not sure if it would be, probably not, but that's not the point). Therefore you torque would be more like 0.375 Nm.

Again I'm not sure about that, but it seems so intuitive.

 

[elbeasto]

Ok thanks. Turns out the part I am looking for is that relationship between the 48v and 9v if any. I know how I'm going to calculate the torque I need, I just need to figure out what is the smallest size battery that can get me there.

Is there some conversion for amp/volt to force generated?

 

[Angello90]

Well I suppose you could use Power = Voltage * Current (P=VI), to calculate the electric power. Than you have mechanical property P(t) = \tau(t) \bullet \omega(t), but I don't know if you can relate these two together. Sorry.

But let's assume that you can than you have:

VI = \tau(t)\bullet\omega(t), knowing your current delivered by a battery you want to use , knowing the torque and knowing the angular speed you can find voltage needed.

Personally I think this would be rather difficult to find as the current drained from the battery would vary depending on the load resistance. Plus Angular velocity would be vary with respect to voltage supplied. It's seems so much simpler to use mains and transformer.

Maybe try adapting batteries from power drills? These are designed fro this sort of a job.

I simply cannot help any more with this sorry. Maybe some other smarter and more experienced person would help you. At this stage I'm out sorry.

Best of luck

 

[elbeasto]

Thanks I appreciate your help anyway!