Does thrust get cancelled by a rotating crank?

[Ripcrow]

I have a frame which supports a crank rotating through 360°. The crank lifts a 30 kg weight 540 times a minute ( 540 rpm). As the crank moves 0° to 90° the weight is accelerated to full speed which means it has thrust but then from 90° to 180° it is slowed to a stop then from 180° to 270° it is accelerated upwards then from 270° to 360° it is decelerated to a stop before commencing the next revolution. My question is will this machine experience force that’s makes it jump up or down. I’m aware that thrust is f=ma and that the weight will have a maximum thrust force at 90° and 270° But how is the deceleration effect of the crank reaching 0° and 180° absorbed. Does the frame need to be weighted down or secured or does a flywheel absorb the thrust during deceleration of the weight.

 

[A.T.]

I’m aware that thrust is f=ma

That's the linear dynamics. With rotating parts you also have angular dynamics. In the general: All the forces on the frame have to balance, and all the moments on the frame have to balance, for the frame to stay static.

 

[Ripcrow]

So the thrust from the linear motion has to be dissipated then. Balancing the rotating parts is easy enough but if the deceleration of the linear motion needs to be absorbed how do I do that.

 

[Filip Larsen]

My question is will this machine experience force that’s makes it jump up or down.

As I understand your description the crank applies lifting force (i.e. work) to the load two times per revolution, i.e. 18 times per second in normal operation, and assume you are then asking to what extend this jerky lifting can be "felt" at the machines ground support points.

Short answer is that it depends on the exact mechanical links between the different parts. For instance, if the crank is suspended relative to the ground support, this suspension most likely will be designed to smooth out some or most of the vibrations, but you would probably still be able to measure the vibration of the lifting force at the ground support.

On the other hand, if we look at the worst-case scenario then it also possible to make some mechanical links that actually amplifies the vibrations, possibly even to the point where the ground support force is positive (i.e. the support legs of the machine accelerate upwards for a short while) or possibly just low enough that any lateral force from the rotating part will allow the support legs to vibrate across a non-level floor.

Which is kind of what happens when you forget to remove the transport bolt from a washing machine :

 

[sophiecentaur]

Is this a pile driver / hammer that's being described? Details would make things more interesting. The radius of rotation is relevant here, to get an idea of the relative accelerations and forces.

So the thrust from the linear motion has to be dissipated then.

As @A.T. comments, the rotational dynamics will also come into it but the range of angles of a long string / push rod can be small so the rotational forces may not be so relevant (Small horizontal components).
The term "dissipated" is not really appropriate for a Force. If there is a thrust force on something then there will be a reaction force back on the force originator. You can balance the force on the crank by having two cranks and weights (as in a multi cylinder IC Engine) or just accept that the crank supporting assembly has to be massive enough (aka 'weighted down') so that very little Energy is transferred to it.

but then from 90° to 180° it is slowed to a stop

If the shaft rotates slowly, the weight force will be providing a force so the crank doesn't need to 'pull the weight back down'. It's a matter of the net vertical acceleration of the weight being downwards all the time. Too fast and the whole assembly could jump off its bed which would then require it to be bolted down.

then from 180° to 270° it is accelerated upwards

If 180 - 270 is the downward phase then I think you mean it is accelerate downwards (by gravity plus the motor torque)

from 270° to 360° it is decelerated to a stop

With an upwards force from the ground.

Does the frame need to be weighted down or secured

Unless there are two flywheels, one each side of the crank, there will be a net torque which will need to be countered by the mounting being fixed to the ground.

 

[Ripcrow]

The difference is the washing machine is unbalanced while my machine is accelerated and decelerated by a crank which should bring a smooth acceleration and deceleration. Absorbing the thrust as it’s decelerated is what has got me confused as maths says it won’t have thrust at 360° but at 270° It will have a large amount of thrust. How does thrust get absorbed to zero by the deceleration of the mass. Does this deceleration cause thrust to transfer from the object being decelerated to the support structure such as bearing and connecting rods. If it is transferred to the supporting structure is it as simple as applying a downward force such as a weight or can something else be done such as rubber mounts or damper springs

 

[Ripcrow]

Is this a pile driver / hammer that's being described? Details would make things more interesting. The radius of rotation is relevant here, to get an idea of the relative accelerations and forces.

As @A.T. comments, the rotational dynamics will also come into it but the range of angles of a long string / push rod can be small so the rotational forces may not be so relevant (Small horizontal components).
The term "dissipated" is not really appropriate for a Force. If there is a thrust force on something then there will be a reaction force back on the force originator. You can balance the force on the crank by having two cranks and weights (as in a multi cylinder IC Engine) or just accept that the crank supporting assembly has to be massive enough (aka 'weighted down') so that very little Energy is transferred to it.

If the shaft rotates slowly, the weight force will be providing a force so the crank doesn't need to 'pull the weight back down'. It's a matter of the net vertical acceleration of the weight being downwards all the time. Too fast and the whole assembly could jump off its bed which would then require it to be bolted down.

If 180 - 270 is the downward phase then I think you mean it is accelerate downwards (by gravity plus the motor torque)

With an upwards force from the ground.Unless there are two flywheels, one each side of the crank, there will be a net torque which will need to be countered by the mounting being fixed to the ground.

It’s a reciprocating saw. It cuts on the down stroke and movement is about 600 mm and weight I’m guessing will be about 30 kg. The acceleration is from 0° to 90°as the crank rotates and then decelerates to 180° before acceleration to 270° before again decelerating to 360°. All this is done 540 times a minute so I’m very concerned about the amount of vibration the saw will produce and even how high it will jump. There is commercially available saws that are as smooth as silk so I’m trying to understand how they smooth out the thrust differences or do I not need to worry about it as the crank slowing the deceleration will make the thrust disappear smoothly

 

[sophiecentaur]

Thanks - that is useful information.

commercially available saws that are as smooth as silk

Yeah - that sort of thing can be very annoying when you realize that 'their' version probably uses much lighter weight parts and smaller bearings. BTW, this sounds like a pretty massive saw if the reciprocating part is 30kG. Could you reduce that at all?
We haven't mentioned damping yet and I would bet that's one way they deal with the vibration. I would imagine that they may select a clever, eccentric flywheel with its mass in anti phase with the load. Some car engines have a separate 'balancing shaft' to deal with vibration but that would probably be too much aggro for a machine tool.
You have a serious problem here in that it calls for quite a bit of 'development' work and experimental mods. From that (scary) video of the suicidal washing machine, it seems that good suspension with damping could be a solution. But the saw table needs to be firmly attached to the bench so that the work is held steady. Perhaps the motor could be allowed to move along the direction of the saw motion - or the drive belt could be at right angles to the saw axis? Then the saw plus saw frame and drive could float. Google images could be a source of ideas.

 

[Ripcrow]

Watch the video to get a better understanding of what I wish to build.

 

[jrmichler]

NOW I understand what you are trying to do. What you have is a crank slider mechanism with 30 kg moving mass, a stroke of 600 mm peak to peak, and running 9 revolutions per second. If the connecting rod is relatively long relative to the stroke, the motion will be approximately sinusoidal.

We start with the equation of motion for the saw: Position = 0.3 meters * sin(9 * 2 * PI * t), where the position is in meters, and t (time) is in seconds. The 2*PI converts revolutions to radians.

Then the velocity is the derivative of position: Velocity = 0.3 meters * 9 * 2 * PI * cos(9 * 2 * PI * t), where velocity is in meters per second.

And the acceleration is the derivative of velocity: Acceleration = 0.3 meters * (9 * 2 * PI)^2 * sin(9 * 2 * PI * t), where acceleration is in meters per second squared. I'm ignoring a negative sign because we are interested only in peak forces.

Since we want the peak acceleration (or velocity), we plug in the angle where the sine or cosine is one, and the peak acceleration is 960 meters per second squared. The peak force is then 960 * 30 kg = 29,000 N. That's peak force up and peak force down.

If the machine weighs less than 2940 kg, it will hop up and down similar to the washing machine. In practice, since the ground under the machine has a spring constant, you need about ten times that mass to keep it from jumping around. That mass is typically a concrete pad. The machine must be firmly bolted to the concrete pad, and the mounting bolts need a safety factor of at least ten.

I have seen machines that broke loose from their foundations. One was a 200 hp reciprocating air compressor, the other was an MG set with an electrical problem. The ten times factor that I recommend is an absolute minimum. Better would be 20 or 30 times.

The bearings and support structure need to be designed for continuous operation at the loads calculated above. The support structure needs to be designed for fully reversed fatigue at those loads.

If you are familiar with trig functions and mass vs force, you should have no trouble following what I did above. You absolutely need to be able to do these calculations yourself, and understand what you are doing, if you expect to make a machine that actually works. If not, you are in danger of making something that disintegrates like that washing machine.

After you understand the loads, you need to design bearings, mounts, support structure, drive mechanism, etc. I recommend that you find a mechanical engineer to help you.

 

[sophiecentaur]

If the machine weighs less than 2940 kg, it will hop up and down similar to the washing machine.

This wouldn't happen if there is a 30kg mass of crank and counterweight, on a damped suspension (as in a motor car?). The two masses will move against each other with no movement of the CM. The only significant force against the frame would be the force from the saw against the wood and that force will depend on the speed of the saw through the wood. It will be zero at each change of direction.
The only disadvantage to this solution that I can see is that you need clearance below the table for the drive to move.

I recommend that you find a mechanical engineer to help you.

Agreed. Any motor with that sort of Power needs to be handled properly.

 

[A.T.]

Any motor with that sort of Power needs to be handled properly.

Especially when it's attached to a saw.

 

[Ripcrow]

NOW I understand what you are trying to do. What you have is a crank slider mechanism with 30 kg moving mass, a stroke of 600 mm peak to peak, and running 9 revolutions per second. If the connecting rod is relatively long relative to the stroke, the motion will be approximately sinusoidal.

We start with the equation of motion for the saw: Position = 0.3 meters * sin(9 * 2 * PI * t), where the position is in meters, and t (time) is in seconds. The 2*PI converts revolutions to radians.

Then the velocity is the derivative of position: Velocity = 0.3 meters * 9 * 2 * PI * cos(9 * 2 * PI * t), where velocity is in meters per second.

And the acceleration is the derivative of velocity: Acceleration = 0.3 meters * (9 * 2 * PI)^2 * sin(9 * 2 * PI * t), where acceleration is in meters per second squared. I'm ignoring a negative sign because we are interested only in peak forces.

Since we want the peak acceleration (or velocity), we plug in the angle where the sine or cosine is one, and the peak acceleration is 960 meters per second squared. The peak force is then 960 * 30 kg = 29,000 N. That's peak force up and peak force down.

If the machine weighs less than 2940 kg, it will hop up and down similar to the washing machine. In practice, since the ground under the machine has a spring constant, you need about ten times that mass to keep it from jumping around. That mass is typically a concrete pad. The machine must be firmly bolted to the concrete pad, and the mounting bolts need a safety factor of at least ten.

I have seen machines that broke loose from their foundations. One was a 200 hp reciprocating air compressor, the other was an MG set with an electrical problem. The ten times factor that I recommend is an absolute minimum. Better would be 20 or 30 times.

The bearings and support structure need to be designed for continuous operation at the loads calculated above. The support structure needs to be designed for fully reversed fatigue at those loads.

If you are familiar with trig functions and mass vs force, you should have no trouble following what I did above. You absolutely need to be able to do these calculations yourself, and understand what you are doing, if you expect to make a machine that actually works. If not, you are in danger of making something that disintegrates like that washing machine.

After you understand the loads, you need to design bearings, mounts, support structure, drive mechanism, etc. I recommend that you find a mechanical engineer to help you.

I’ll have to do some of those equations so I understand it correctly. Seems it’s a bigger undertaking then I thought. Thanks for your assistance

 

[sophiecentaur]

Seems it’s a bigger undertaking then I thought.

If you could get a glimpse (photo) of what lies under the saw table. Does it move or is it fixed to the frame? How is the drive delivered to the saw frame?

 

[sophiecentaur]

I found this Russian Patent which seems to have some relevant discussion. They have definitely tried to deal with your problem!
See what you think.

 

[Ripcrow]

The ones I have seen all use a crank arrangement and arms connecting to the saw. Is it possible that a top mounted crank with a weight suspended from it would smooth out the thrust. For example if the saw unit ( the linear moving part of the saw ) weighed 30 kg could a 30 kg weight be set up to work in the opposite direction. So as the crank lifts the saw unit the weights are going down and when the saw is on its way down the crank is lifting the weights. To my thinking this would balance the thrust generated. Am I right in assuming this ?

 

[sophiecentaur]

To my thinking this would balance the thrust generated. Am I right in assuming this ?

I think so. If the system acts horizontally, it's easier to think about but the principle would be right for vertical operation - it would just need some suspension to take up the vertical oscillations. Without a workpiece for the saw to cut, the two halves of the system would balance each other out and there would be zero / small force on the support. The crank and saw frame would move from side to side but their centre of mass would not move if the 30kg balancing mass on the crank is opposite to the big end bearing.
As soon as the wood is introduced, there will be a force on the saw which will act on the support (action and reaction). The force on the saw will be a maximum when it is moving fastest through the wood (I think) which would be when the oscillatory forces on the reciprocating mass will be minimal. I think this arrangement would mean that the forces against the suspension would be greatest when the saw is working hardest and go to (nearly) zero with no load.

top mounted crank with a weight suspended from it

The crank would have a 300mm radius of sweep? The 600mm of motion seems enormous but I guess it's all on a big scale and you need to get rid of the chips on each stroke. The balance mass could be over or under the saw table but I think it could be safer underneath. Also the motor etc would be out of the way; but you would be better qualified to have an opinion on this.
I was originally thinking in terms of a heavy off centre weight on the crank but that wouldn't be as good as your suspended mass idea because there would be no lateral forces. I think this would involve a double crank. (no problem) I guess the length of the top and bottom push rods would need to be the same to keep the relative angles and forces the same over the whole cycle. This is very much along the lines of a Flat IC engine and there's loads written about them.