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Default Power/Force of hydraulic cylinder???

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


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Default Power/Force of hydraulic cylinder???

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.
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"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


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Default Power/Force of hydraulic cylinder???

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
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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


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"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


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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


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Default Power/Force of hydraulic cylinder???

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.
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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
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Default Power/Force of hydraulic cylinder???

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.

--
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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


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Default Power/Force of hydraulic cylinder???

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.


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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.
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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
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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
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On Jan 16, 6:37*pm, wrote:
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.


He is correct. The force that counts is that applied at the load.
That depends on all the various levers, angles, etc. but it still
comes out to "what is the actual vertical force on the load". It will
be considerably less than the force measured 'at the cylinder pivots'.

Harry K
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On Jan 16, 11:38*pm, Harry K wrote:
On Jan 16, 6:37*pm, wrote:

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


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.


He is correct. *The force that counts is that applied at the load.
That depends on all the various levers, angles, etc. but it still
comes out to "what is the actual vertical force on the load". *It will
be considerably less than the force measured 'at the cylinder pivots'.


I also read this statement to suggest that he was thinking in terms of
vector mechanics such that only the vertical component of the
cylinder's force vector was applied to the load. If so. this has very
little to do with this type of system. The cylinder could be
horizontal, sideways of pointing down and still supply (almost)
exactly the same force at the load.

On second reading, it's really hard to tell exactly what he meant by
that second clause...


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Larry Fishel wrote:
On Jan 16, 11:38 pm, Harry K wrote:
On Jan 16, 6:37 pm, wrote:

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


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.

He is correct. The force that counts is that applied at the load.
That depends on all the various levers, angles, etc. but it still
comes out to "what is the actual vertical force on the load". It will
be considerably less than the force measured 'at the cylinder pivots'.


I also read this statement to suggest that he was thinking in terms of
vector mechanics such that only the vertical component of the
cylinder's force vector was applied to the load. If so. this has very
little to do with this type of system. The cylinder could be
horizontal, sideways of pointing down and still supply (almost)
exactly the same force at the load.

On second reading, it's really hard to tell exactly what he meant by
that second clause...


Well, all of the above...

The poster who started this subthread to which I was responding noted
that altho had sizable cylinder on a tractor loader that didn't have
anything close to the lift force that one might expect simply from the
bore/pressure of the cylinder and the responder directly ahead mentioned
only the weight of the loader itself as a confounding factor...

A typical small/medium tractor loader is generally a very simple
mechanism a la the one shown at the link below--

http://www.americanlisted.com/idaho_12/garden_house_22/ford_8_n_loader_tractor_3375_14793104.html

While there are complex mechanisms found on either very large and/or
expensive loaders and/or the compact loaders such as the Bobcat and ilk,
a basic loader of the above type doesn't have mechanical advantage--the
upward component at the bottom is fairly small and the relatively long
moment arm beyond the lift point (necessary to get reasonable lift
height, etc.) requires quite a bit to counteract.

I was simply trying to be concise at, perhaps, the cost of some clarity
in pointing out that mechanics are a part as well as weight and cylinder
force...as per usual on usenet, the opportunity to seek advantage and
display perceived intellectual prowess at expense of others is,
apparently, irresistible force for some.

--
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dpb wrote:
....

A typical small/medium tractor loader is generally a very simple
mechanism ...


OK, I went and dug up the operator's manual off the JD148 that's on the
4440 as the loader tractor here...

http://www.tractordata.com/farm-tractors/000/0/9/96-john-deere-4440-photos.html

It's hydraulics are 2250 psi, lift cylinders are 2.5" bore/1.75" rod --
from which one can estimate max lift force at the lift pin of sotoo
11000 lbf. The published breakout lift force is "only" 4700 lbf in
comparison. It'll handle 2000-lb round bales two-high loading semi's
easily, however as far as the weight; wish the reach were about 6" more,
though, as one has to be sure to have the fork as near the bottom as
possible and if they're just a little loosely wrapped they can rotate
some so maneuvering can sometimes slow one down loading. When there are
nearly a thousand to go; any time wasted, even a few seconds/bale, is
sorely trying to the soul...

I'll try to post somewhere a picture from last summer's haying altho
don't think I stopped to take pictures while actually loading out; had
my hands full of controls at that point...

--
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dpb wrote:
....
OK, I went and dug up the operator's manual off the JD148 that's on the
4440 as the loader tractor here...

....
It's hydraulics are 2250 psi, lift cylinders are 2.5" bore/1.75" rod --
from which one can estimate max lift force at the lift pin of sotoo
11000 lbf. The published breakout lift force is "only" 4700 lbf in
comparison. ...


OK, just for grins on a cold, damp day...

I went out and measured the hinge point distances and the overall length
of the moment arm to the end of the lift arms (accounting for the angle,
measuring to the extended straight line from the fixed pin thru the
upper cylinder lifting point).

When down at ground level, the pertinent dimensions and angles are --

Well, let's see if I can do enough ASCII art to get a picture as crude
as it will be for referencing to...


Arm pin (A) Cylinder Pin (U)
X --------------X--------O End actuator attach pin (E)
|
|
--X Lower cylinder pin (L)

The (nearly) closed cylinder length at ground level (L-U) is 42"

Solving the small right triangle to the left of L we find the angle from
the horizontal to the line A-L is 75.5deg. Using the law of cosines
the angle ALE is 96.4. The cylinder lift angle is the complement of
these two or 180-(75.5+96.4) = 8.1deg.

The vertical component at U is then 11000*sin(8.1) == 1550 lb

Balancing the moment I got that the lift at U was 2200 lb, within 10% of
the 2350-lb breakaway spec (remember this is looking at only the one
cylinder).

FWIW (which ain't a lot, but it was cold, damp and blustery enough
yesterday I didn't try to do much outside).

Unfortunately, the photo bucket site didn't let me log on yesterday and
I couldn't get a flickr acc't to go thru either so gave up on the
posting of the pictures for the moment...

--

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"dpb" wrote in message
...
dpb wrote:
...
OK, I went and dug up the operator's manual off the JD148 that's on the
4440 as the loader tractor here...

...
It's hydraulics are 2250 psi, lift cylinders are 2.5" bore/1.75" rod --
from which one can estimate max lift force at the lift pin of sotoo 11000
lbf. The published breakout lift force is "only" 4700 lbf in comparison.
...


OK, just for grins on a cold, damp day...

I went out and measured the hinge point distances and the overall length
of the moment arm to the end of the lift arms (accounting for the angle,
measuring to the extended straight line from the fixed pin thru the
upper cylinder lifting point).

When down at ground level, the pertinent dimensions and angles are --

Well, let's see if I can do enough ASCII art to get a picture as crude
as it will be for referencing to...


Arm pin (A) Cylinder Pin (U)
X --------------X--------O End actuator attach pin (E)
|
|
--X Lower cylinder pin (L)

The (nearly) closed cylinder length at ground level (L-U) is 42"

Solving the small right triangle to the left of L we find the angle from
the horizontal to the line A-L is 75.5deg. Using the law of cosines
the angle ALE is 96.4. The cylinder lift angle is the complement of
these two or 180-(75.5+96.4) = 8.1deg.

The vertical component at U is then 11000*sin(8.1) == 1550 lb

Balancing the moment I got that the lift at U was 2200 lb, within 10% of
the 2350-lb breakaway spec (remember this is looking at only the one
cylinder).

FWIW (which ain't a lot, but it was cold, damp and blustery enough
yesterday I didn't try to do much outside).

Unfortunately, the photo bucket site didn't let me log on yesterday and I
couldn't get a flickr acc't to go thru either so gave up on the posting of
the pictures for the moment...

--
At this point this thread is kind of stale so I don't know how many people
will take notice. Despite the fact that just about everything you did went
way over my head, just want to say that I'm wicked impressed





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RBM wrote:
....

... Despite the fact that just about everything you did went
way over my head, ...


Chuckle...mine, too!!!

The earlier question got me wondering about what was the vertical
component in actual fact -- I was surprised to realize the lift
cylinders are only 8 degrees above horizontal when the bucket is on the
ground.

If it hadn't been such an ugly day out I'm sure I wouldn't have actually
done it...

--


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dpb wrote:
dpb wrote:

....
Well, let's see if I can do enough ASCII art to get a picture as crude
as it will be for referencing to...

Arm pin (A) Cylinder Pin (U)
X --------------X--------O End actuator attach pin (E)
|
|
--X Lower cylinder pin (L)

....
the angle ALE is 96.4. The cylinder lift angle is the complement of
these two or 180-(75.5+96.4) = 8.1deg.


.... the angle ALU is 96.4. ...

....
... the lift at U was 2200 lb, ...


.... the lift at _E_ ...

...gave up on the posting of the pictures for the moment...


....

Coupla' typos noted above and here are a coupla' pictures took while
"gathering up" in preparation of loadout. First is just shot of bales
as rolled; second shows a couple of piles. Each group is a stack of 34,
what'll go on a flatbed so can load at one time instead of having to go
gather up while loading. The last is the home place; can just barely
see the house thru the cedars/and two largest elms...a little white
siding shows through.

http://picasaweb.google.com/104782889132331211356/UntitledAlbum?authkey=Gv1sRgCNbg6vPHmO6C1gE&feat=d irectlink#5565465227185160130

--
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"dpb" wrote in message
...
dpb wrote:
dpb wrote:

...
Well, let's see if I can do enough ASCII art to get a picture as crude
as it will be for referencing to...

Arm pin (A) Cylinder Pin (U)
X --------------X--------O End actuator attach pin (E)
|
|
--X Lower cylinder pin (L)

...
the angle ALE is 96.4. The cylinder lift angle is the complement of
these two or 180-(75.5+96.4) = 8.1deg.


... the angle ALU is 96.4. ...

...
... the lift at U was 2200 lb, ...


... the lift at _E_ ...

...gave up on the posting of the pictures for the moment...


...

Coupla' typos noted above and here are a coupla' pictures took while
"gathering up" in preparation of loadout. First is just shot of bales as
rolled; second shows a couple of piles. Each group is a stack of 34,
what'll go on a flatbed so can load at one time instead of having to go
gather up while loading. The last is the home place; can just barely see
the house thru the cedars/and two largest elms...a little white siding
shows through.

http://picasaweb.google.com/104782889132331211356/UntitledAlbum?authkey=Gv1sRgCNbg6vPHmO6C1gE&feat=d irectlink#5565465227185160130

--That is some beautiful, wide open space. Where is it, and where's the
snow?



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Default Power/Force of hydraulic cylinder???

RBM wrote:
"dpb" wrote in message

....
... and here are a coupla' pictures took while
"gathering up" in preparation of loadout. ...

http://picasaweb.google.com/104782889132331211356/UntitledAlbum?authkey=Gv1sRgCNbg6vPHmO6C1gE&feat=d irectlink#5565465227185160130

--That is some beautiful, wide open space. Where is it, and where's the
snow?

....



Located far SW KS, just about 50 mi from CO right on OK line.

It is pretty country when it rains; we'd had a nice one in late August
and another in early in Sept about two weeks before those were taken so
stuff had had time to green up nicely after swathing. I hadn't realized
the date wasn't set on the camera and intended to reset it but forgot
about it. Those were taken about 10 Sept last fall.

Unfortunately, that was also about the last moisture we've had since;
it's serious dry at the moment. The winter wheat is in poor shape where
got it in and up at all.

We've missed out on all the snows except for just a couple that gave us
only a dusting to a short inch each; all the actual moisture has been
north/east or south. That's a fairly typical problem out here
particularly in the La Nina pattern that tends to form a ridge of high
pressure over the western High Plains that pushes the jet stream/storm
track around to the east/north.

--
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