Calculating Required Torque for 12V DC Motor

• howardt0818
In summary, the conversation discusses the design of a reciprocating saw that will use a scotch-yoke mechanism to convert rotational kinetic energy into linear kinetic energy. The final step in the design process is to calculate the required torque for a 12V motor that will be used to power the saw. The saw blade weight, desired cut travel, and desired RPM are also mentioned as important factors to consider. The equation for calculating the required torque is discussed, with the additional factors of power, current, voltage, and type of motor being taken into account. The conversation also touches on the importance of understanding the force applied by the saw teeth, the type of material being cut, and the tooth profile. The use of different saw-teeth styles
howardt0818
TL;DR Summary
I am trying to calculate the required Torque to drive an object at a given speed.
I am designing a reciprocating saw that I plan to use a scotch-yoke mechanism to convert rotational kinetic energy into linear kinetic energy.
I want to spec out a motor and the final thing I need to calculate is the required torque. The motor must be 12 Volts and I need to know the required torque so I can purchase a motor.

Desired cut travel: 2 inches
Desired RPM: 800 RPM

To achieve a cut travel of 2 inches, the scotch yoke circle that will be attached directly to the motor must be 2 inches in diameter.

What equation could I use to calculate this?

Power and current can be altered to achieve this but the voltage will be operating at 12V and the motor will be DC geared most-likely.

If I calculate the load torque as simply the force multiplied by the radius - 1.5lb-in in this case - does that mean this is also the torque rating required by the motor to achieve 700 RPM?

Welcome to PF.
You will need to know the force applied by the teeth as they cut the (as yet) unspecified material, and how many teeth will be engaged at one time.
That will depend on the tooth profile, and the method used to cut or rip saw dust crumbs out of the bulk.

Lnewqban
howardt0818 said:
If I calculate the load torque as simply the force multiplied by the radius - 1.5lb-in in this case - does that mean this is also the torque rating required by the motor to achieve 700 RPM?
Consider that your motor will need to accelerate and decelerate that mass between zero and maximum cutting speed in a sinusoidal way twice the rotational speed.
The energy required to cut also depends on the normal force that the tool applies onto the material and on the type of cooling-lubricant to be used, if any.

The mechanism itself wil steal some energy as well due to friction (which increases with cuting normal force) and vibrations.
You will need an additional slider located closer to the motor, as wel as a mechanism to follow the depth of the cut while keeping steady normal force.

Last edited:
@howardt0818
Different materials are cut with different styles of saw-teeth.

A bread-knife is designed to slice, without producing crumbs. The teeth vary the pressure applied to the cutting surface. But a bread-knife is not really a saw that cuts; a knife slices.

A chainsaw or a circular saw, collects the saw-dust cut in a large gullet during the pass of the stroke, until the dust is carried from the slot being cut = kerf, and then released from the gullet outside the kerf. That does not work well with a reciprocating saw as the gullets get clogged.

A reciprocating hacksaw has many small teeth that alternately cut, and progressively pass the crumbs along and out of the kerf. Each tooth is offset to the side to cut a wider kerf than the blade material. The hacksaw blade should be lifted while it is being returned for the next stroke. That reduces tooth wear and pushes the dust from the kerf more efficiently.

Your diagram seems to show a 2” tooth pitch, with a 2” reciprocation. That will not function well as a saw because there may be parts which do not get cut, while material may be trapped in the kerf. But it might work as a bread-knife, or for slicing soft material.

That is why the material and the tooth geometry are important.
What will you be cutting.

berkeman and Lnewqban
What are you cutting? 800 SPM is way too fast for cutting steel if you want any blade life at all. That blade speed will cut wood just fine, but the mechanical forces at 13 cuts per second will be challenging. The Scotch yoke needs clearance to operate, and the load rattles back and forth within that clearance. Too much clearance will cause it to pound itself to pieces. Too little clearance will cause binding. Scotch yokes work best in an oil bath because the oil helps to cushion the impact of the load moving back and forth within the clearance.

How large a motor? Start by measuring the cutting force by pulling a blade with a spring scale. Then multiply by about three to account for different materials and blades, add in the slider bearing friction, and add in the Scotch yoke friction. Then multiply by the one inch radius to get driven torque. Add in the speed reducer friction and divide by the speed reducer ratio to get the motor torque.

The cutting force is not the mass of the blade. The cutting force can be much larger than the total weight of the blade, especially with a new, sharp blade. And depending on the material being cut.

Lnewqban
Baluncore said:
Welcome to PF.
You will need to know the force applied by the teeth as they cut the (as yet) unspecified material, and how many teeth will be engaged at one time.
That will depend on the tooth profile, and the method used to cut or rip saw dust crumbs out of the bulk.
This will be cutting leafy greens, 30 inches of saw blade and 30 inch bed of leafy greens. For simplicity sake let's just focus on the weigh of the blade and less so on the actual cutting, as I am mostly trying to calculate specs for the motor to simply move the saw blade at a desired speed.

jrmichler said:
What are you cutting? 800 SPM is way too fast for cutting steel if you want any blade life at all. That blade speed will cut wood just fine, but the mechanical forces at 13 cuts per second will be challenging. The Scotch yoke needs clearance to operate, and the load rattles back and forth within that clearance. Too much clearance will cause it to pound itself to pieces. Too little clearance will cause binding. Scotch yokes work best in an oil bath because the oil helps to cushion the impact of the load moving back and forth within the clearance.

How large a motor? Start by measuring the cutting force by pulling a blade with a spring scale. Then multiply by about three to account for different materials and blades, add in the slider bearing friction, and add in the Scotch yoke friction. Then multiply by the one inch radius to get driven torque. Add in the speed reducer friction and divide by the speed reducer ratio to get the motor torque.

The cutting force is not the mass of the blade. The cutting force can be much larger than the total weight of the blade, especially with a new, sharp blade. And depending on the material being cut.
Apologies for leaving those fine details out. This will be for a leafy-green plant harvesting mechanism. The motor will be operated at 12V - size, current, and power can be decided after the fact. This is mostly risk reduction protocol I am taking at the moment to see how well the scotch-yoke mechanism can work for exactly this purpose. If successful, I can then get into the finer details such as force-per-tooth. Thanks!

Baluncore said:
@howardt0818
Different materials are cut with different styles of saw-teeth.

A bread-knife is designed to slice, without producing crumbs. The teeth vary the pressure applied to the cutting surface. But a bread-knife is not really a saw that cuts; a knife slices.

A chainsaw or a circular saw, collects the saw-dust cut in a large gullet during the pass of the stroke, until the dust is carried from the slot being cut = kerf, and then released from the gullet outside the kerf. That does not work well with a reciprocating saw as the gullets get clogged.

A reciprocating hacksaw has many small teeth that alternately cut, and progressively pass the crumbs along and out of the kerf. Each tooth is offset to the side to cut a wider kerf than the blade material. The hacksaw blade should be lifted while it is being returned for the next stroke. That reduces tooth wear and pushes the dust from the kerf more efficiently.

Your diagram seems to show a 2” tooth pitch, with a 2” reciprocation. That will not function well as a saw because there may be parts which do not get cut, while material may be trapped in the kerf. But it might work as a bread-knife, or for slicing soft material.

That is why the material and the tooth geometry are important.
What will you be cutting.
Apologies for leaving out that information. This will be for cutting leafy-greens grown on a 30-inch bed.

It seems that you are slicing, not sawing.
I assume you cut down against the leafy material, held against a cutting board.
There should be no material produced as dust.
The cut material will fall away from the knife.
Is that correct?

Baluncore said:
It seems that you are slicing, not sawing.
I assume you cut down against the leafy material, held against a cutting board.
There should be no material produced as dust.
The cut material will fall away from the knife.
Is that correct?
There will not be a cutting board. There will be a rotating "mop-head" like system to sweep the cut leaves behind the saw and onto a conveyer belt. If you would like, I can provide images of a similar product that resembles the action I am trying to accomplish.

A picture would be useful, as a link or drag and drop onto your next post.

Baluncore said:
A picture would be useful, as a link or drag and drop onto your next post.

I see. I believe you show what is often called a finger mower.
It is usual to have a fixed plate with fingers that supports the reciprocating cutting knife. The fixed comb of fingers prevent the crop being moved sideways by the knife. The reciprocating knife is made from many triangular teeth that slice the crop against the sides of the fingers.
That suggests your knife will be reciprocating against a part lubricated plate, rather than in free space, so there will be additional friction in the system.

Because of the weight of the reciprocating knife, you will need an eccentric counterbalance on the drive shaft, or it will shake apart the motor and the bearings. The weight of the circulating counterbalance will change the peak torque requirement on the motor.
I cannot see how you can balance that easily with a scotch yoke driving the knife.

Baluncore said:
I see. I believe you show what is often called a finger mower.
It is usual to have a fixed plate with fingers that supports the reciprocating cutting knife. The fixed comb of fingers prevent the crop being moved sideways by the knife. The reciprocating knife is made from many triangular teeth that slice the crop against the sides of the fingers.
That suggests your knife will be reciprocating against a part lubricated plate, rather than in free space, so there will be additional friction in the system.

Because of the weight of the reciprocating knife, you will need an eccentric counterbalance on the drive shaft, or it will shake apart the motor and the bearings. The weight of the circulating counterbalance will change the peak torque requirement on the motor.
I cannot see how you can balance that easily with a scotch yoke driving the knife.
If not a scotch yoke, what would be a typical method of driving a machine like this?

howardt0818 said:
If not a scotch yoke, what would be a typical method of driving a machine like this?
For a vertical axis motor, a crank pin with half the balance weight above and half below the connecting rod to the knife.
If using a finger mower, or sickle mower, with a cutter bar you might search for “balanced cutter bar”, or “cutter bar” with “double alternate movement”. They do not have the same balance problem. They are more like a hand held hedge trimmer.
The style of teeth on the bar will be determined by the width of the leaf or stem you are cutting.

Baluncore said:
For a vertical axis motor, a crank pin with half the balance weight above and half below the connecting rod to the knife.
If using a finger mower, or sickle mower, with a cutter bar you might search for “balanced cutter bar”, or “cutter bar” with “double alternate movement”. They do not have the same balance problem. They are more like a hand held hedge trimmer.
The style of teeth on the bar will be determined by the width of the leaf or stem you are cutting.
On a more simplified matter - in order for the motor to rotate the 1.5 pound load attached to it, assume a maximum of 2 lbs including friction forces, how would one calculate the torque required to rotate said load at that speed? EDIT: Power can be up to 40 Watts available

howardt0818 said:
how would one calculate the torque required to rotate said load at that speed?
It goes something like this.
Convert everything to SI units. Then it all comes out in watts = joule per second.

A force of 2 pounds ≈ 1 kg force = 9.8 Newton. ( kgf * g = force in Newton ).
Because you reciprocate through 2”, the radius of the crank is 1” minimum = 25.4 mm.
Final drive shaft torque = 9.8 N * 0.025 m = 0.245 N·m.

There are 2π radians per turn, and 60 sec in a minute.
750 RPM = 750 / 60 = 12.5 rev/sec = 12.5 * 2π = 78.54 radians/sec .

Power in watts = torque * angular velocity.
∴ Required power = 0.245 N·m * 78.54 radians/sec = 19.25 watts.

1. How do I calculate the required torque for a 12V DC motor?

To calculate the required torque for a 12V DC motor, you will need to know the motor's speed, power consumption, and efficiency. The formula for calculating torque is torque = (power / angular velocity) x (60 / 2π). This will give you the torque in Nm (Newton-meters).

2. What is the relationship between voltage and torque in a DC motor?

The relationship between voltage and torque in a DC motor is directly proportional. This means that as the voltage increases, the torque will also increase. This is because a higher voltage will result in a stronger magnetic field, which will produce more torque.

3. How does the load affect the required torque for a 12V DC motor?

The load will directly affect the required torque for a 12V DC motor. The heavier the load, the more torque will be required to move it. This is because the motor will need to work harder to overcome the resistance of the load.

4. Can the required torque for a 12V DC motor be too high?

Yes, the required torque for a 12V DC motor can be too high. If the torque required is too high, it can cause the motor to overheat and potentially damage it. It is important to calculate the required torque accurately to ensure the motor can handle the load.

5. Are there any other factors that can affect the required torque for a 12V DC motor?

Yes, there are other factors that can affect the required torque for a 12V DC motor. These include friction, air resistance, and the efficiency of the motor. It is important to consider all of these factors when calculating the required torque to ensure the motor can perform the necessary task efficiently and safely.

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