I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated

RBM wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated


The things that matter are the PSI and the diameter of the bore.
Figure the surface area of the internal diameter of the cylinder.
3.14 X half the bore X half the bore. Multiply that by the PSI.
I come up with 14,836#. That is, of course, under ideal circumstances.
I think the old hydraulic control valves used to have a pressure
relief set at 2000 psi.

"Dean Hoffman" wrote in message
...
RBM wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I
can figure from information derived from the net, the bore would be 3".
It has a stroke of 18". It will be fed from a pump delivering 8 GPM @
2100 PSI.

Any help would be appreciated


The things that matter are the PSI and the diameter of the bore.
Figure the surface area of the internal diameter of the cylinder.
3.14 X half the bore X half the bore. Multiply that by the PSI.
I come up with 14,836#. That is, of course, under ideal circumstances.
I think the old hydraulic control valves used to have a pressure relief
set at 2000 psi.

OK, that actually sounds reasonable at something around 7 tons. This was on
a log splitter with a 5 HP engine, but I have no idea what it's pump put
out, but it's a really small pump. I'm rearranging it to go on a small skid
steer with auxiliary hydraulics.

Thanks

RBM wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated



I screwed up in my previous answer. I figured only part of the
problem. This will give you a better answer.
http://tinyurl.com/4slybla

On Jan 16, 12:23*pm, Dean Hoffman wrote:
RBM wrote:

I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated

* * I screwed up in my previous answer. *I figured only part of the
problem. *This will give you a better answer.
* *http://tinyurl.com/4slybla

14,844 from the calculator vs. 14,836 for your # crunching
Yeah, be more careful next time - you were off by .05%

R

"RicodJour" wrote in message
...
On Jan 16, 12:23 pm, Dean Hoffman wrote:
RBM wrote:

I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I
can
figure from information derived from the net, the bore would be 3". It
has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated

I screwed up in my previous answer. I figured only part of the
problem. This will give you a better answer.
http://tinyurl.com/4slybla

14,844 from the calculator vs. 14,836 for your # crunching
Yeah, be more careful next time - you were off by .05%

R

Close enough for Govment work

That's about the sternest rebuke I've heard on the group, in
ages. That's got to hurt. NOT!

--
Christopher A. Young
Learn more about Jesus
www.lds.org
..


"RicodJour" wrote in message
...
I screwed up in my previous answer. I figured only part of
the
problem. This will give you a better answer.
http://tinyurl.com/4slybla

14,844 from the calculator vs. 14,836 for your # crunching
Yeah, be more careful next time - you were off by .05%

R

On Sun, 16 Jan 2011 11:30:21 -0500, "RBM" wrote:

I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated

Pressure pushing or pulling? If pushing the shaft diameter does not
have any bearing on the force delivered. The gallons per minute
likewize have no effect on the force delivered 3 inch bore is 7 square
inches of area, so the push force delivered is 7X2100= 14,700 lbs
force.

If it is a double acting cyl, the pull-back area is 7-3.14=3.86 sq
inches, and the pull-back force is 3.86X2100=8906 lbs force.

On Jan 16, 12:46*pm, wrote:
On Sun, 16 Jan 2011 11:30:21 -0500, "RBM" wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated

*Pressure pushing or pulling? If pushing the shaft diameter does not
have any bearing on the force delivered. The gallons per minute
likewize have no effect on the force delivered 3 inch bore is 7 square
inches of area, so the push force delivered is 7X2100= 14,700 lbs
force.

If it is a double acting cyl, the pull-back area is 7-3.14=3.86 sq
inches, and the pull-back force is 3.86X2100=8906 lbs force.

Correct. To get a bit pedantic: The 'power' is usually limited by
the valve. When it comes to logsplitters the valve is usually set at
2,500 - 2,750 psi. The next class action suit should be for the
manufacturers of log splitters. Their force claims (usually 22 ton
for a 3" cylinder) are way over inflated.

Harry K

In article , RBM wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated



If your bore diameter and shaft diameters are correct, the force on the
piston side of the cylinder will be about 14800 pounds, or 7.4 tons. If
it is a double acting cylinder, the force on the rod side will be about
8240 pounds, or 4.1 tons. Hope this is not a homework problem.




--
Often wrong, never in doubt.

Larry Wasserman - Baltimore Maryland - lwasserm(a)sdf. lonestar. org

On Sun, 16 Jan 2011 23:40:21 +0000 (UTC),
(Larry W) wrote:

In article , RBM wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated



If your bore diameter and shaft diameters are correct, the force on the
piston side of the cylinder will be about 14800 pounds, or 7.4 tons. If
it is a double acting cylinder, the force on the rod side will be about
8240 pounds, or 4.1 tons. Hope this is not a homework problem.

I have a loader on my tractor with two fairly large cylinders. I can
lift a maximum of 800lbs, or lift a 1000lb hay bale a couple inches
off the ground, but that's all.

wrote in message
news
On Sun, 16 Jan 2011 23:40:21 +0000 (UTC),
(Larry W) wrote:

In article , RBM wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I
can
figure from information derived from the net, the bore would be 3". It
has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated



If your bore diameter and shaft diameters are correct, the force on the
piston side of the cylinder will be about 14800 pounds, or 7.4 tons. If
it is a double acting cylinder, the force on the rod side will be about
8240 pounds, or 4.1 tons. Hope this is not a homework problem.

I have a loader on my tractor with two fairly large cylinders. I can
lift a maximum of 800lbs, or lift a 1000lb hay bale a couple inches
off the ground, but that's all.

I had a small compact Ford tractor. The cylinders on the loader were
considerably smaller that this, which has a 3.5" diameter. Mine could only
pick up about 600 pounds. You also have to keep in mind that your cylinders
aren't only picking up the hay bale, they're also picking up the weight of
the bucket and arms of the loader.

RBM wrote:
....

I had a small compact Ford tractor. The cylinders on the loader were
considerably smaller that this, which has a 3.5" diameter. Mine could only
pick up about 600 pounds. You also have to keep in mind that your cylinders
aren't only picking up the hay bale, they're also picking up the weight of
the bucket and arms of the loader.

They're also not pointing vertically directly below the load, either;
only the force in the direction of lift is actually vertical.

--

On Sun, 16 Jan 2011 19:16:03 -0600, dpb wrote:

RBM wrote:
...

I had a small compact Ford tractor. The cylinders on the loader were
considerably smaller that this, which has a 3.5" diameter. Mine could only
pick up about 600 pounds. You also have to keep in mind that your cylinders
aren't only picking up the hay bale, they're also picking up the weight of
the bucket and arms of the loader.

They're also not pointing vertically directly below the load, either;
only the force in the direction of lift is actually vertical.
Not true. Not at all. You need to study your basic physics - simple
levers, classes and ratios.

On Sun, 16 Jan 2011 19:56:20 -0500, "RBM" wrote:


wrote in message
news
On Sun, 16 Jan 2011 23:40:21 +0000 (UTC),
(Larry W) wrote:

In article , RBM wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I
can
figure from information derived from the net, the bore would be 3". It
has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated



If your bore diameter and shaft diameters are correct, the force on the
piston side of the cylinder will be about 14800 pounds, or 7.4 tons. If
it is a double acting cylinder, the force on the rod side will be about
8240 pounds, or 4.1 tons. Hope this is not a homework problem.

I have a loader on my tractor with two fairly large cylinders. I can
lift a maximum of 800lbs, or lift a 1000lb hay bale a couple inches
off the ground, but that's all.

I had a small compact Ford tractor. The cylinders on the loader were
considerably smaller that this, which has a 3.5" diameter. Mine could only
pick up about 600 pounds. You also have to keep in mind that your cylinders
aren't only picking up the hay bale, they're also picking up the weight of
the bucket and arms of the loader.


And the cyl might have a 18" stroke, and the loader might lift the
bucket 180 inches. This means, forgetting the weight of the arms and
buckets for a minute, you would only have a 1480 lb lifting capacity.
10:1 lift ratio means 1:10 load ratio.

On Sun, 16 Jan 2011 18:04:22 -0600, wrote:

On Sun, 16 Jan 2011 23:40:21 +0000 (UTC),
(Larry W) wrote:

In article , RBM wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated



If your bore diameter and shaft diameters are correct, the force on the
piston side of the cylinder will be about 14800 pounds, or 7.4 tons. If
it is a double acting cylinder, the force on the rod side will be about
8240 pounds, or 4.1 tons. Hope this is not a homework problem.

I have a loader on my tractor with two fairly large cylinders. I can
lift a maximum of 800lbs, or lift a 1000lb hay bale a couple inches
off the ground, but that's all.
That is due to the leverage. You have a "negative" ratio.

In article ,
wrote:
On Sun, 16 Jan 2011 23:40:21 +0000 (UTC),
(Larry W) wrote:

In article , RBM wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated



If your bore diameter and shaft diameters are correct, the force on the
piston side of the cylinder will be about 14800 pounds, or 7.4 tons. If
it is a double acting cylinder, the force on the rod side will be about
8240 pounds, or 4.1 tons. Hope this is not a homework problem.

I have a loader on my tractor with two fairly large cylinders. I can
lift a maximum of 800lbs, or lift a 1000lb hay bale a couple inches
off the ground, but that's all.


I'm not sure what point you are trying to make, if any. In a typical
loader design, the cylinders act through a lever an fulcrum with
a negative mechanical force advantage. Even if the cylinder develops
a force of say 10 tons, the force available at the end of the loader
arms will be significantly less. Usually the geometry of a loader
or backhoe design has the maximum mechanical advantage available in
curling the bucket.



--
Better to be stuck up in a tree than tied to one.

Larry Wasserman - Baltimore Maryland - lwasserm(a)sdf. lonestar.org

wrote in message
...
On Sun, 16 Jan 2011 11:30:21 -0500, "RBM" wrote:

I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has
a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated


You have the force answer, if your pump really puts out 8 GPM at 2100
PSI it will fill that cylinder in around 2 seconds.

Is that a Campbell Hausfeld pump? We know how they lie ;-)

It's the auxiliary hydraulic pump built into a Thomas 85 skid steer. Thomas
was built in Canada, don't know who's pump they use, but I really doubt it's
a crappy C/H

wrote:

On Sun, 16 Jan 2011 11:30:21 -0500, "RBM" wrote:

I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I can
figure from information derived from the net, the bore would be 3". It has a
stroke of 18". It will be fed from a pump delivering 8 GPM @ 2100 PSI.

Any help would be appreciated


You have the force answer, if your pump really puts out 8 GPM at 2100
PSI it will fill that cylinder in around 2 seconds.

What this means is that the splitter should function well with the skid
steer at or near idle speed.