Can a Hydraulic Motor Determine Torque and RPM for a Two Chain Sprocket System?

[dakotahm88]

TL;DR

Two sprocket chain system help!

[Mentor Note -- postprocessed attached image for clarity]

Hello!

Looking for a formula(s) on how I can determine the Torque and RPM needed on the chain (by hydraulic motor) to move the load in the illustration at 240 inches per minute.
The load will vary, so I won’t ever really need to push or pull 20,000lbs at that speed, but I do need to be able to reach that speed in times of no load.
The skate that Carries the load is fixed to an I-beam with cam followers to help reduce friction between the skate and beam it slides on.

 

[berkeman]

Welcome to PF.

Summary: Two sprocket chain system help!

Looking for a formula(s) on how I can determine the Torque and RPM needed on the chain (by hydraulic motor) to move the load in the illustration at 240 inches per minute.
The load will vary, so I won’t ever really need to push or pull 20,000lbs at that speed, but I do need to be able to reach that speed in times of no load.

As you probably know, the torque requirements depend on the acceleration of the system (and the mass of the system) and the friction components in the system. I don't think you've specified either yet...

 

[Lnewqban]

Welcome!
Is that thing vertical?
10 tons?
No counter-weight?

 

[jack action]

For the RPM requirement, it is based on your desired velocity $v$ and the sprocket diameter $d$:
$$RPM = \frac{v\text{ [in/min]}}{3.1416 \times d \text{ [in]}}$$
Or:
$$RPM = \frac{240\text{ in/min}}{3.1416 \times 10.410 \text{ in}} = 7.34 \text{ rpm}$$

Whenever you will determine the desired load $F$ that needs to move at your desired speed $v$, the horsepower required by your motor will be:
$$HP = \frac{F \text{ [lb]} \times v \text{ [in/min]}}{396\ 000 \text{ lb.in/min/hp}}$$
Or, in your case:
$$HP = \frac{F \text{ [lb]} \times 240 \text{ in/min}}{396\ 000 \text{ lb.in/min/hp}} = \frac{F \text{ [lb]}}{1650 \text{ lb/hp}}$$
So for every 1650 lb of load, you will need 1 hp to move it at 240 in/min. There will most likely be some losses in transmission of the power that you will need to take into account as well.

Once you will adjust the motor RPM to the desired sprocket RPM (7.34 rpm) - with a gear reducer or something similar - the torque will automatically match your needs, as long as the motor produces the required horsepower at that RPM.

 

[dakotahm88]

Welcome to PF.As you probably know, the torque requirements depend on the acceleration of the system (and the mass of the system) and the friction components in the system. I don't think you've specified either yet...

I don’t care how fast it accelerates only that it’s capable of said speed at smaller loads.
The only friction, that I know of, is where the 8 cam followers run against the i-beam. And yes, vertical.

 

[berkeman]

And yes, vertical.

Vertical?! Holy smokes. With no counterweight? That increases the size of the required motor many times. What is the application? Is it an elevator? Does it involve life safety? What is your level of education in Mechanical Engineering?

 

[dakotahm88]

Vertical?! Holy smokes. With no counterweight? That increases the size of the required motor many times. What is the application? Is it an elevator? Does it involve life safety? What is your level of education in Mechanical Engineering?

It’s to upgrade a water well drilling rig, it doesn’t actually carry a load of 20,000lbs. It pushes the bit into the ground with the capability of 20,000 lbs. Realistically with pressure reliefs it will be set to rarely exceed about 1800 lbs. It just needs to be rated for 20,000, incase a bit gets stuck and we need to pull harder to get it out.

We build derricks on well service rigs that handle up to 250,000 LBs with 6 line blocks.

Just new to the angular stuff.

We originally were going to use a ball screw, but decided against it, what’s cool, is the horsepower calculation for the ball screw came out to be almost identical.

I have a degree in Petroleum Engineering. Took a lot of classes for mechanical just for fun.

 

[dakotahm88]

Vertical?! Holy smokes. With no counterweight? That increases the size of the required motor many times. What is the application? Is it an elevator? Does it involve life safety? What is your level of education in Mechanical Engineering?

It’s to upgrade a water well drilling rig, it doesn’t actually carry a load of 20,000lbs. It pushes the bit into the ground with the capability of 20,000 lbs. Realistically with pressure reliefs it will be set to rarely exceed about 1800 lbs. It just needs to be rated for 20,000, incase a bit gets stuck and we need to pull harder to get it out.

We build derricks on well service rigs that handle up to 250,000 LBs with 6 line blocks.

Just new to the angular stuff.

We originally were going to use a ball screw, but decided against it, what’s cool, is the horsepower calculation for the ball screw came out to be almost identical.

I have a degree in Petroleum Engineering. Took a lot of classes for mechanical just for fun

 

[dakotahm88]

Not sure why my photos are blurry. Anyways, issue we are having is getting the speed slow enough to use it. Low RPM high torque pumps are like 240 RPM, and we were installing without gear reduction. Even with a flow controller set as low as possibly it’s still to fast.

 

[Averagesupernova]

That 1.75 inch shaft seems small to be pushing down with 22,000 lbs.

 

[dakotahm88]

That 1.75 inch shaft seems small to be pushing down with 22,000 lbs.

It’s only 4” wide, fixed-on both ends to bearings.
A clevis with a safe working load of 30 tons is only hung with a 1-3/4” pins

Pretty sure it will work.

 

[Baluncore]

Anyways, issue we are having is getting the speed slow enough to use it. Low RPM high torque pumps are like 240 RPM, and we were installing without gear reduction. Even with a flow controller set as low as possibly it’s still to fast.

I would select the hydraulic motor based on the available hydraulic flow and the pump pressure relief valve setting. If the volumetric capacity of the motor is greater than necessary, then the motor will turn slower with all the flow, so no gear reduction is needed.

You know the safe maximum load on the chain. Once you pick a motor capacity, you should protect the chain by limiting the maximum motor hydraulic pressure. Acceleration is then not a problem, the motor will accelerate at the maximum possible rate with the torque below the set limit, as the over-pressure bypass valve across the motor will lift.

For the existing hydraulic system, that will later drive the motor, I have not seen the maximum pump flow rate, nor the pump relief valve pressure setting.

 

[dakotahm88]

I would select the hydraulic motor based on the available hydraulic flow and the pump pressure relief valve setting. If the volumetric capacity of the motor is greater than necessary, then the motor will turn slower with all the flow, so no gear reduction is needed.

You know the safe maximum load on the chain. Once you pick a motor capacity, you should protect the chain by limiting the maximum motor hydraulic pressure. Acceleration is then not a problem, the motor will accelerate at the maximum possible rate with the torque below the set limit, as the over-pressure bypass valve across the motor will lift.

For the existing hydraulic system, that will later drive the motor, I have not seen the maximum pump flow rate, nor the pump relief valve pressure setting.

Relief setting will depend which pump we decide to go with. Was looking at a 19.36 Cu in./rev pump that would operate at a maximum rating of 2031 PSI. 3450 in/lbs torque. I believe is roughly 11 HP.
190 RPM rated at 15.9 GPM.

11 HP x 1650 lbs = max pull of 18,150 lbs
Chain is rated at 15,700 lbs. So we would adjust the relief pressure to lower the maximum torque to not exceed that. Average Breaking strength of the chain is like 108,000 Lbs, not sure why they use such a large safety factor.

 

[Baluncore]

Was looking at a 19.36 Cu in./rev pump that would operate at a maximum rating of 2031 PSI. 3450 in/lbs torque. I believe is roughly 11 HP.
190 RPM rated at 15.9 GPM.

Convert all to SI = MKSA.
19.36 Cu in./rev @ 190 RPM. Flow = 19.36 * 190 = 3678.4 cu in /min.
3678.4 / 60 = 61.3 in3 /sec = 1.00 litre /sec ;
Pressure 2031 psi = 140 bar = 14000 kPa = 14 MPa ;
Power W = pressure Pa * volume m3 = 14e6 * 1e-3 = 14 kW.
14 kW = 18.77 HP, (assumes 100% efficiency).
I wonder why that is that different to your 11 HP.
Are you sure that pump is 190 RPM and not 1900 RPM, more like a 110 HP diesel engine ?
It makes a factor of 10 difference.

The 17 tooth sprocket; 10.41” diameter. Circumference = 10.41 * Pi = 32.70”
The chain pitch of 32.70” / 17 = 1.92”, is non-standard = suspicious.
140-2 ASA is a duplex roller chain, with a pitch of 1-3/4".
The circumference of the 17 tooth sprocket is therefore 1.75” * 17 = 29.75”.
Please verify the pitch of the chain and the number of teeth.

The maximum speed of 240”/min will require 240” / 29.75” = 8.07 RPM.
At full flow, 8.07 RPM requires a motor capacity of 19.36 cu” * 190 / 8.07 = 455.8 cu”/rev.
455.8 cu" = 7469 cc/rev = 7.5 litre. And yes, it will have a very thick shaft.

I believe that is an uneconomic solution, so it is time to stop, check, and think.
I would consider a flow divider or a mechanical RPM reduction.

 

[Tom.G]

Average Breaking strength of the chain is like 108,000 Lbs, not sure why they use such a large safety factor.

Perhaps due to the "fun" of retrieving it from the well bottom?

 

[dakotahm88]

Convert all to SI = MKSA.
19.36 Cu in./rev @ 190 RPM. Flow = 19.36 * 190 = 3678.4 cu in /min.
3678.4 / 60 = 61.3 in3 /sec = 1.00 litre /sec ;
Pressure 2031 psi = 140 bar = 14000 kPa = 14 MPa ;
Power W = pressure Pa * volume m3 = 14e6 * 1e-3 = 14 kW.
14 kW = 18.77 HP, (assumes 100% efficiency).
I wonder why that is that different to your 11 HP.
Are you sure that pump is 190 RPM and not 1900 RPM, more like a 110 HP diesel engine ?
It makes a factor of 10 difference.

The 17 tooth sprocket; 10.41” diameter. Circumference = 10.41 * Pi = 32.70”
The chain pitch of 32.70” / 17 = 1.92”, is non-standard = suspicious.
140-2 ASA is a duplex roller chain, with a pitch of 1-3/4".
The circumference of the 17 tooth sprocket is therefore 1.75” * 17 = 29.75”.
Please verify the pitch of the chain and the number of teeth.

The maximum speed of 240”/min will require 240” / 29.75” = 8.07 RPM.
At full flow, 8.07 RPM requires a motor capacity of 19.36 cu” * 190 / 8.07 = 455.8 cu”/rev.
455.8 cu" = 7469 cc/rev = 7.5 litre. And yes, it will have a very thick shaft.

I believe that is an uneconomic solution, so it is time to stop, check, and think.
I would consider a flow divider or a mechanical RPM reduction.

The sprocket measurement I posted is to the top of the teeth, your math is bottom of the teeth, which is more accurate, but I didn’t have that measurement. Yes, pitch is 1.75”.

You figured the pumps peak int capacity.
See attachment. I went by its maximum rated continuous use. I’d rather have a little too much then not enough. I can always easily limit flow and pressure to the motor.

And yes it’s 190RPm cont., 1” shaft.

 

[dakotahm88]

Sorry forgot attachment.

 

[dakotahm88]

Convert all to SI = MKSA.
19.36 Cu in./rev @ 190 RPM. Flow = 19.36 * 190 = 3678.4 cu in /min.
3678.4 / 60 = 61.3 in3 /sec = 1.00 litre /sec ;
Pressure 2031 psi = 140 bar = 14000 kPa = 14 MPa ;
Power W = pressure Pa * volume m3 = 14e6 * 1e-3 = 14 kW.
14 kW = 18.77 HP, (assumes 100% efficiency).
I wonder why that is that different to your 11 HP.
Are you sure that pump is 190 RPM and not 1900 RPM, more like a 110 HP diesel engine ?
It makes a factor of 10 difference.

The 17 tooth sprocket; 10.41” diameter. Circumference = 10.41 * Pi = 32.70”
The chain pitch of 32.70” / 17 = 1.92”, is non-standard = suspicious.
140-2 ASA is a duplex roller chain, with a pitch of 1-3/4".
The circumference of the 17 tooth sprocket is therefore 1.75” * 17 = 29.75”.
Please verify the pitch of the chain and the number of teeth.

The maximum speed of 240”/min will require 240” / 29.75” = 8.07 RPM.
At full flow, 8.07 RPM requires a motor capacity of 19.36 cu” * 190 / 8.07 = 455.8 cu”/rev.
455.8 cu" = 7469 cc/rev = 7.5 litre. And yes, it will have a very thick shaft.

I believe that is an uneconomic solution, so it is time to stop, check, and think.
I would consider a flow divider or a mechanical RPM reduction.

“The maximum speed of 240”/min will require 240” / 29.75” = 8.07 RPM.
At full flow, 8.07 RPM requires a motor capacity of 19.36 cu” * 190 / 8.07 = 455.8 cu”/rev.
455.8 cu" = 7469 cc/rev = 7.5 litre. And yes, it will have a very thick shaft.”

So your saying this motor is way too small?
I was going by the other guys 1HP per 1650 Lbs.

I was just thinking, if I used a flow regulator and limited the flow to the pump prior to the bank to about 80% of the rated flow, and then used a variable flow friction detent valve with a relief set at about 1600 PSI, that I could barely open the valve, allowing it to turn much more slowly but with 1800 psi, it would still be capable of the 15,400 lbs at a much slower speed then it’s maximum rating. Guess I was not thinking correctly.

 

[dakotahm88]

I guess what I’m thinking. There will be times when your going back up to latch another drill stem that moving a few feet per second would be better. I just need the capability of crawling. If this don’t work I need some better suggestions.

 

[Baluncore]

I was designing based on constant flow hydraulics. That will give the best economy, and minimise heat generated in the hydraulic fluid.
I think the problem may be with the slow, high volume pump selected.
Is that pump present for some other purpose, or can another be selected?
What is the source of motive power, can it be greater than 190 RPM?
What is the maximum chain speed when moving drill stem ? 2 ft/sec = 1440”/min ?
What is the minimum steady chain speed when pulling the drill string ? 240”/min ?
The speed ratio there would be 1440 / 240 = 6.0
Maybe the shaft could be driven by one of two different motors, depending on the required task.
An alternative would be a variable flow system.

 

[dakotahm88]

I was designing based on constant flow hydraulics. That will give the best economy, and minimise heat generated in the hydraulic fluid.
I think the problem may be with the slow, high volume pump selected.
Is that pump present for some other purpose, or can another be selected?
What is the source of motive power, can it be greater than 190 RPM?
What is the maximum chain speed when moving drill stem ? 2 ft/sec = 1440”/min ?
What is the minimum steady chain speed when pulling the drill string ? 240”/min ?
The speed ratio there would be 1440 / 240 = 6.0
Maybe the shaft could be driven by one of two different motors, depending on the required task.
An alternative would be a variable flow system.

Variable flow would be easier then two different motors in this situation.
Two feet per second would be perfect.
And yes 240”/min while drilling.
I just chose that pump because that’s the slowest pump I could find with required torque. I have not purchased it yet, so I’m up for using anything.

 

[Baluncore]

I just chose that pump because that’s the slowest pump I could find with required torque.

What is the source of motive power to drive the pump ?
Did you plan to use gear reduction to drive the pump ?

A diesel engine or an electric motor would directly spin the pump at about 1500 RPM, so the pump could be reduced in capacity to 19.36 * 190 / 1500 = 2.45 cu"/rev. That would provide the same flow rate and power.

The smaller pump would cost less and will eliminate the need for gear reduction.

 

[dakotahm88]

What is the source of motive power to drive the pump ?
Did you plan to use gear reduction to drive the pump ?

A diesel engine or an electric motor would directly spin the pump at about 1500 RPM, so the pump could be reduced in capacity to 19.36 * 190 / 1500 = 2.45 cu"/rev. That would provide the same flow rate and power.

The smaller pump would cost less and will eliminate the need for gear reduction.

The pump is PTO driven off truck 1:1. So yeah ideally would run about 1500 RPM. I can put any size we want, there’s three pumps in total for other stuff on the truck.
And several more motors.

 

[dakotahm88]

The pump is PTO driven off truck 1:1. So yeah ideally would run about 1500 RPM. I can put any size we want, there’s three pumps in total for other stuff on the truck.
And several more motors.

I had No intentions of using gear reduction due to limited space

 

[berkeman]

It’s to upgrade a water well drilling rig, it doesn’t actually carry a load of 20,000lbs. It pushes the bit into the ground with the capability of 20,000 lbs. Realistically with pressure reliefs it will be set to rarely exceed about 1800 lbs. It just needs to be rated for 20,000, incase a bit gets stuck and we need to pull harder to get it out.

We build derricks on well service rigs that handle up to 250,000 LBs with 6 line blocks.

Just new to the angular stuff.

We originally were going to use a ball screw, but decided against it, what’s cool, is the horsepower calculation for the ball screw came out to be almost identical.

I have a degree in Petroleum Engineering. Took a lot of classes for mechanical just for fun.

Thread will remain closed.

@dakotahm88 -- I will be contacting your company's insurance company to be sure that they are okay with you getting your design advice off of the Internet instead of actually getting your Professional Engineer certificate to do this kind of work. Have a nice day.