Hydraulic Drive Question 1

[spd748]

I am in the process of designing a hydraulic drive for a circular saw mill. I have most of the numbers worked out except one. I can't seem to calculate the torque required to move the carriage. The specs are as follows:

I need to move a 5000 lb (maximum total mass)carriage along a flat plane at a maximum velocity of 325 fpm. The carriage rides on steel roller bearing wheels which in turn ride on a steel track so friction should be minimal. The carriage is pulled by a cable which is wrapped around a 13" steel drum. The drum is mounted on a shaft which will be turned by a hydraulic motor. I would like to keep acceleration/de-acceleration rates to a minumum to allow the system to start and more importantly stop the carriage quickly. I ran the specs through a couple of formulas that I have but the torque requirement that I come up with sounds way too low. I suppose that I need an independent opinion on the matter. Thanks in advance for any information that respondents may offer.

 

[spd748]

Sorry, I left a few things out. The input shaft of the winch drum needs to turn ~95 rpm to pull the carriage at the required 325 fpm. The carriage is pulled along the track which moves the log or cant through the head saw. The hydraulic drive motor is then reversed to pull the carriage back to the starting point. Another reason to keep acceleration/de-acceleration rates to a minumum is to speed up the production of the mill as the carriage must start, stop, start and stop to complete one cycle of operation. Thanks again.

 

[spd748]

I read my posts and realized that I still haven't asked a clear question. I am trying to calculate the amount of torque that needs to be applied to the input shaft of the 13" diameter winch drum in order to move the carriage. One thing that I can't (or don't know how to) calculate is the amount of resistance that will occur once the log or cant hits the saw itself. Thoughts?

 

[hydtools]

How quickly do you want to get to speed? Or what is the travel distance to reach speed? How heavy is the log or is that weight included in the 5000lbs ? How quickly do you want to stop or reverse the travel? Or what is the travel distance to stop/reverse the travel direction? How heavy is the steel drum?

Answers to these questions will determine torque required to bring the load to speed and then bring it to a stop and then reverse travel.

Ted

 

[desertfox]

Hi spd748

What about the inertia of the drum and as the cable winds round the drum the torque will change have you considered this as well?

desertfox

 

[spd748]

Thanks for the reply Ted. The 5000 lbs is total mass. Actually its a bit over the maximum mass. I added 20% to the weight of the carriage and the heaviest log that my mill will handle as a safety factor. The distance from the leading edge of the longest log that my mill will handle to the leading edge of the saw is ~60". This distance is measured on a log that is 20' long. The headsaw is centered on a 50' track with the maximum travel distance of the carriage being 35'. Once the trailing edge of the longest log leaves the trailing edge of the saw (the cut is complete) the carriage can travel an additional ~60" before reaching its travel limit. I haven't weighed the winch shaft/drum. I would estimate it to be in the area of 200lbs.

 

[spd748]

desertfox,

The cable wraps the drum 7 times forming a single layer. There is no change in diameter as the cable feeds on/off the drum. I did not calculate the inertia of the drum. Good point.

 

[desertfox]

Hi spd748

I see a bit of a problem here if you have only 60" to spare before the log hits the saw then your normal running speed of 325fpm means the carriage would cover that distance in less than 1 second so if you need to accelerate the carriage upto running speed then unless I have misunderstood something your going to have to do it pretty fast.

Prior to the above had you been able to run up to speed in say 2.5 secounds according to my calcs using a rolling resistance of 0.006 then just get the carriage moving without acceleration your looking at 16.25lbft, with the acceleration of 2.16ft/s^2 (based on 2.5 seconds) I get a torque figure of 182lbft so adding the two together I get just under 200lbft to start and get your carriage upto speed.

As regards energy required to the carriage during sawing I was thinking along the lines of an energy balance ie kinetic energy of carriage against the workdone in cutting the log and convert this back to additional torque that your motor needs.

desertfox

 

[EdDanzer]

The force to push the log through the saw can be very high. I would suggest looking at what other manufactures use for carriage drive horsepower.

http://www.machinerysales.com/equipment.php?cid=10

will provide some information. By using the speed and the hp you can get a torque value.
Back in the early 1980’s I worked in a smaller system and I think it had over 25 HP driving the carriage and it was marginal in a 24” green fir log.

Ed Danzer

http://www.danzcoinc.com

http://www.dehyds.com

 

[spd748]

desertfox,

Are you saying that the 200 lb-ft need to be applied to the winch drum shaft, meaning your calculations take into account the mechanical disadvantage of the drum radius?

Another note, how much torque would be required to accelerate the carriage to speed in the 60" clearance?

Ed,

I called several manufacturers. None of them seemed to be able to tell me what size (cu. in./rev) hydraulic motor they used in their drives while all of them had price and shipping information. Go figure. I did find some older mills on the web which have been converted to hydraulic drive. I emailed each of them but haven't received any replies.

From my research, a hydraulic motor capable of 200 lb-ft is not all that big. I attempted to compare the prices that I was given by the manufacturers that I called to the prices of motors at several hydraulic suppliers. If price is any indication as to motor size then the motors that are being used in modern hydraulic mills are massive, on the order of 20k in-lbs of output.

 

[desertfox]

Hi spd748

Not sure what you mean by mechanical disadvantage I just worked out what the straight pull force the carriage would need then multiplied it by the drum radius.
It doesn't take into account other friction or efficiency losses,inertia's (like the drum inertia).
What torque figures did you arrive at?

According to my calculations to be upto speed in half a second would take about 11,127.72 lbin or 927.31 lbft which again doesn't allow for the losses mentioned above.

The final torque you need of course must include the energy needed for cutting the wood.


desertfox

 

[spd748]

desertfox,

You covered my question in your last post. The disadvantage I was speaking of was the radius of the drum, i.e., you multiplied the straight line pull by the drum radius.

During normal sawing operations, it is rare that I would ever drive the carriage up to full operating speed before the blade hits the wood. If fact, I usually ease into the wood until the saw clears the top of the log, then drive it home. The 325 fpm mentioned is the fastest the carriage would ever need to travel as the maximum "bite" of the blade is .11" per tooth per revolution. I have a 55 tooth blade so .11" x 55 teeth x 650 rpm = 3932.5 ipm or 327.7 fpm. In the real world, it is nearly impossible to drive the carriage at that speed continiously through the cut.

 

[desertfox]

hi spd748

So how do the torque figures compare with your calculations?

Okay so thats a maximum speed you need, so it might not be to bad after all.
Whats the configuration like of the saw relative to the log, I envisage a twenty foot log heading toward a saw blade and the saw blade diameter being vertical as you look from the side and basically the saw splits the log into two
halfs down its twenty foot length is that correct?

desertfox

 

[EdDanzer]

The smallest Tyrone carriage drive at Machinery Sales is 40 hp. 40 hp at 95 rpm generates 26,537 in lbs torque. I would suggest using a piston motor in a planetary drive for the best performance. If you choose a low speed high torque motor use some sort of overhung load adaptor or a shaft coupler between it and the drive pulley.

Ed Danzer

http://www.danzcoinc.com

http://www.dehyds.com

 

[spd748]

I came up with 660 lbft to drive the carriage to 325 fpm in 1 second. This type of calculation is new to me so I didn't feel all that confident in my product.

The saw is configured as you envision. A log is loaded on the carriage and dogged or clamped in place. The carriage is then pulled along the track frame toward the headsaw. The headsaw splits a measured amount from the outside edge of the log. The carriage is then pulled back along the track frame to the starting point. The carriage is then indexed to whatever the next board thickness is going to be and the cycle begins again.

Twenty feet is the maximum length my mill will handle. The average length I am cutting is twelve feet.

I plan on installing a pressure compensated flow control valve between the spool and motor. This way I can positively control the maximum motor speed. I'll install a pump that will produce sufficient volume to drive the motor/carriage at ~325 fpm. The flow control valve can be adjusted to compensate for changes in conditions, knotty logs, harder woods, etc with 'on the fly' fine adjustments being made at the spool.

After looking over your calculations, it appeares that a Charlynn 4000 series 19 cu in/rev motor will work. This motor produces 3820 lb-in @ 300 rpm on 1500 psi and 25 gpm. With a gear reduction of 3:1, that should provide 100 rpm and 11460 lb-in.

Thoughts?

 

[desertfox]

Hi spd748

Whoa there my 11,127lbin is only for starting and moving the carriage, won't you need some more torque for the carriage while sawing the log and also what about drum intertia, friction,efficiency etc.
Unless you ahead of me lol, I was looking into the sawing of wood see what I could find.
Your torque figure about right for one second, it depends on how much rolling friction you assume you have.

desertfox

 

[EdDanzer]

I’m sure you want to save money on components costs, but be careful as it will cost much more to do it twice.

A hydrostatic drive will generate much less heat, provide better control and last longer than what you are thinking. Flow controls are heat generators in this type of application. It is better to oversize the components that have to replace them when it does not work.

Ed Danzer

http://www.danzcoinc.com

http://www.dehyds.com

 

[spd748]

You are correct. I haven't settled on that motor yet. I just looked at a few to see what was available based on the torque requirements so far. Sorry for the scare.

I'm still at a loss as to the method of calculating the additional torque required while the saw is in the cut. I tried every search engine I know of and several reference books looking for something on this subject. I found an estimate indicating that each tooth on the saw required approximately 3 horsepower to pull it through the cut. This figure was found in a section of a reference book on how to choose engine power when setting up a mill. If I understand correctly, this is the power required to move the teeth through the wood. As such, I don't see where this figure helps me in determining the resistance of the cutting action as it relates to carriage movement.

I understand that the torque required to get the carriage moving will be significantly greater than the torque required to keep it moving thus there will be an amount of torque available once the wood hits the blade.

The current design of the mill uses flat belts to drive the carriage. There are a set of 2 of these belts that wrap around a pulley which drives the cable drum. One belt is used for each direction of travel. Perhaps the photo will better explain this.

 

[spd748]

Ed,

I agree. I want to keep costs at a minimum. I also don't want to start replacing parts after a few hours or after the system doesn't work at all. In the past I've found that it's better to think (or dream) of something at least twice as long as I ever plan on using it before I spend the first dime. I was given a quote of $6750 for the required parts to convert my mill. In the end, I may end up going this route. I'm still holding out hope that I can design a system that will work at a reduced cost. The thought sounds simple at first. All I need to do is add some type of hydraulic or hydrostatic motor to turn a shaft in two directions. Thats where the simplicity ends.

I have build several very simple hydraulic systems in the past. I wouldn't know where to start on a hydrostatic system.

 

[EdDanzer]

The $6750 may not be a bad price depending on what it includes.
How much do you plan on using the mill? The picture shows a very old system so high production high reliability may not be required. If you are interested in used or surplus parts to build it with contact me outside the forum.

Ed Danzer

http://www.danzcoinc.com

http://www.dehyds.com