How much force does a common shop vise develop in its jaws? Google was
pretty unhelpful on this - the only reference I could find suggested
up to 7000 lb. Another paper suggested that a minimum required
clamping force for machining is much less - up to 1000 N.

I am interested because I am trying to guess how much force I can
develop in this press:

http://www.flickr.com/photos/2768312...57622484352534

Right now my answer would be "not enough" (see the following photos if
you are interested in the process).

Can anybody suggest how such force could be measured?

Michael Koblic,
Campbell River, BC

wrote:
How much force does a common shop vise develop in its jaws? Google was
pretty unhelpful on this - the only reference I could find suggested
up to 7000 lb. Another paper suggested that a minimum required
clamping force for machining is much less - up to 1000 N.

I am interested because I am trying to guess how much force I can
develop in this press:

http://www.flickr.com/photos/2768312...57622484352534

Right now my answer would be "not enough" (see the following photos if
you are interested in the process).

Can anybody suggest how such force could be measured?

Michael Koblic,
Campbell River, BC

Looks a lot like a wine press. I know the small wine press I have will
develop about 1000 pounds at the plate with a 2' lever on the handle.

--
Steve W.

Can anybody suggest how such force could be measured?

Michael Koblic,
Campbell River, BC

Michael

Although not sure of the span dimension I am almost sure I could
easily bend th 1/2" plate so any measurements would be questionable
unless the bend of the span were taken into account.

And from many years of using presses in connection with photography I
am not sure the heat requirements are at all dependent on the
pressure. The pressure is probably needed to assure enough contact so
the heat can do its thing for the mechanical connection of the media.
The heat is probably needed to fuse the media and not completely
dependent on pressure.

My two cents.
Bob AZ

On Thu, 14 Jul 2011 22:37:28 -0700, Bob AZ wrote:
On Thu, 14 Jul 2011 21:13:28 -0700, Michael Koblic wrote:

How much force does a common shop vise develop [...]
http://www.flickr.com/photos/27683124@N07/5938730984/in/set-72157622484352534
Can anybody suggest how such force could be measured?
....
Although not sure of the span dimension I am almost sure I could easily
bend the 1/2" plate so any measurements would be questionable unless the
bend of the span were taken into account.

He used 1/4" plate, and in later picture says bending was a problem.
I expect that 3/4"-1" plate probably is needed to avoid much bending.

And from many years of using presses in connection with photography I am
not sure the heat requirements are at all dependent on the pressure.
The pressure is probably needed to assure enough contact so the heat can
do its thing for the mechanical connection of the media. The heat is
probably needed to fuse the media and not completely dependent on
pressure.

Probably so. Unfortunately, the http://www.techniks.com/how_to.htm
Press-n-Peel web page is a bit vague on requirements, merely saying
"Temperature setting on the iron is critical, [...] Suggested starting
temperature is 275-325 degrees F. [...] Iron until board has completely
and fully reached the temperature of the iron. Time varies with the size
and thickness of the board. Generally this is 1.5 to 10 min." They do
offer a HIX Corporation Heat Transfer Press, on their information page.

--
jiw

On Jul 15, 12:13*am, wrote:
How much force does a common shop vise develop in its jaws?

That's sort of like asking how fast a common car goes... Different
vices have different thread pitch, different handle lengths, and
different (stronger/weaker) humans operating them. So, unless an
answer like "somewhere between 500 and 20,000 pounds" satisfies your
curiosity, you may be disappointed.

You should be able to calculate (roughly) the force your press will
generate using the thread pitch, length of the handles and how much
force YOU can apply to the handles (which may be the hardest to
measure)...assuming it's strong enough to hold together.

wrote in message
news
How much force does a common shop vise develop in its jaws? Google was
pretty unhelpful on this - the only reference I could find suggested
up to 7000 lb. Another paper suggested that a minimum required
clamping force for machining is much less - up to 1000 N.

I am interested because I am trying to guess how much force I can
develop in this press:

http://www.flickr.com/photos/2768312...57622484352534

Right now my answer would be "not enough" (see the following photos if
you are interested in the process).

Can anybody suggest how such force could be measured?

Michael Koblic,
Campbell River, BC


The formula to find the force exerted by your screw is
F= 2*Pi*r*p*f

F is the force of the screw
Pi is 3.14159
r is the distance from the center to the handle
p is the screw pitch
f is the force you exert on the handle

from the photo, I estimate the screw multiplies your force by a factor of
75.

However, evenly applied force is not the way to remove bubbles. You will
just trap them and compress them. The way to remove bubbles would be to
run the plate and film between two rubber rollers with a rigidly-held, but
adjustable gap. Another way is to apply the force through a slightly domed
rubber pad so the center contacts first and the air is pushed to the
outside. You have to make sure the rubber is compressible enough and the
force is great enough that you make contact on the outer edge.

On 15/07/2011 05:13, wrote:
How much force does a common shop vise develop in its jaws? Google was
pretty unhelpful on this - the only reference I could find suggested
up to 7000 lb. Another paper suggested that a minimum required
clamping force for machining is much less - up to 1000 N.

I am interested because I am trying to guess how much force I can
develop in this press:

http://www.flickr.com/photos/2768312...57622484352534

Right now my answer would be "not enough" (see the following photos if
you are interested in the process).

Can anybody suggest how such force could be measured?

Michael Koblic,
Campbell River, BC
It's easy to work out.
Let P = the pitch of the screw, and H = the length of the handle (this
is measured at the point where you grip the handle and is generally a
little less than the actual length).
The mechanical advantage of the system is then 2*Pi*H/P

Assuming the pitch is, say, 0.1", and the handle is 12" long, then the
mechanical advantage is 2*Pi*12/0.1 = 75.4/0.1 = 754
If you apply a force of 10Lb to the end of the handle, then the screw
will exert a force of 7540 Lb, assuming no losses due to friction.
Unfortunately, the frictional losses will be rather high, probably more
than 50%, so the actual force would be more like 3000 Lb.
IHTH

--
Regards, Gary Wooding
(To reply by email, change gug to goog in my address)

On Jul 15, 12:13*am, wrote:
How much force does a common shop vise develop in its jaws? Google was
pretty unhelpful on this - the only reference I could find suggested
up to 7000 lb. Another paper suggested that a minimum required
clamping force for machining is much less - up to 1000 N.

I am interested because I am trying to guess how much force I can
develop in this press:

http://www.flickr.com/photos/2768312.../set-721576224...

Right now my answer would be "not enough" (see the following photos if
you are interested in the process).

Can anybody suggest how such force could be measured?

Michael Koblic,
Campbell River, BC

I just checked the bench and milling vises with a hydraulic load cell,
which is a cylinder of 1 sq inch area attached to a 10,000 PSI gauge.
At a 'reasonable' handle force without hammering the 4" milling vise
reached 1500 lbs, the 3-1/2" bench vise 2000.

jsw

On Fri, 15 Jul 2011 04:06:38 -0700 (PDT), Jim Wilkins
wrote:

On Jul 15, 12:13*am, wrote:
How much force does a common shop vise develop in its jaws? Google was
pretty unhelpful on this - the only reference I could find suggested
up to 7000 lb. Another paper suggested that a minimum required
clamping force for machining is much less - up to 1000 N.

I am interested because I am trying to guess how much force I can
develop in this press:

http://www.flickr.com/photos/2768312.../set-721576224...

Right now my answer would be "not enough" (see the following photos if
you are interested in the process).

Can anybody suggest how such force could be measured?

Michael Koblic,
Campbell River, BC

I just checked the bench and milling vises with a hydraulic load cell,
which is a cylinder of 1 sq inch area attached to a 10,000 PSI gauge.
At a 'reasonable' handle force without hammering the 4" milling vise
reached 1500 lbs, the 3-1/2" bench vise 2000.

Don't you have a 5 or 6" mechanic's shop vise, too, Jim?

I'd be willing to bet that one could do the 7k# force Mikey suggested.
Wilton vises have 25kpsi castings. Another ad suggests 30kpsi.
Tormach uses 80kpsi ductile iron bodies. Neither suggests jaw force.

Cool! I'd never seen an air over hyd vise before.
http://www.posilock.com/Hydraulic%20Vise/Vise.htm

--
Learning to ignore things is one of the great paths to inner peace.
-- Robert J. Sawyer

On Jul 15, 8:51*am, Larry Jaques
wrote:
...
At a 'reasonable' handle force without hammering the 4" milling vise
reached 1500 lbs, the 3-1/2" bench vise 2000.

Don't you have a 5 or 6" mechanic's shop vise, too, Jim?
...

I did but never used it and eventually traded it for a welding table
etc. I can accomplish more through subterfuge than brute force.

The 4" brake 2805-0105 shown here bends thicker stock more neatly:
http://tool.wttool.com/search?w=4%22+brake&x=35&y=9

jsw

On Fri, 15 Jul 2011 06:17:09 -0700 (PDT), Jim Wilkins
wrote:

On Jul 15, 8:51*am, Larry Jaques
wrote:
...
At a 'reasonable' handle force without hammering the 4" milling vise
reached 1500 lbs, the 3-1/2" bench vise 2000.

Don't you have a 5 or 6" mechanic's shop vise, too, Jim?
...

I did but never used it and eventually traded it for a welding table
etc. I can accomplish more through subterfuge than brute force.

Age and treachery will always overcome youth and skill.
(I know. I owned the t-shirt.)


The 4" brake 2805-0105 shown here bends thicker stock more neatly:
http://tool.wttool.com/search?w=4%22+brake&x=35&y=9

http://goo.gl/yEznb I have this type, plus a shorty bender from HF.

--
Learning to ignore things is one of the great paths to inner peace.
-- Robert J. Sawyer

On Thu, 14 Jul 2011 21:13:28 -0700, wrote:

How much force does a common shop vise develop in its jaws? Google was
pretty unhelpful on this - the only reference I could find suggested
up to 7000 lb. Another paper suggested that a minimum required
clamping force for machining is much less - up to 1000 N.

I am interested because I am trying to guess how much force I can
develop in this press:

http://www.flickr.com/photos/2768312...57622484352534

Right now my answer would be "not enough" (see the following photos if
you are interested in the process).

Can anybody suggest how such force could be measured?

Michael Koblic,
Campbell River, BC

Use a trailer tougue scale or a Snap on brake force gage. I happen to
have a snap-on brake force gauge set, they are only 1/2" thick and
made to go in a caliper in place of the brake pads to check caliper
pressure. Measure 0-5000 pounds.

http://www.sherline.com/lm.htm



Remove 333 to reply.
Randy

On Fri, 15 Jul 2011 08:50:03 -0400, Randy333
wrote:

On Thu, 14 Jul 2011 21:13:28 -0700, wrote:

How much force does a common shop vise develop in its jaws? Google was
pretty unhelpful on this - the only reference I could find suggested
up to 7000 lb. Another paper suggested that a minimum required
clamping force for machining is much less - up to 1000 N.

I am interested because I am trying to guess how much force I can
develop in this press:

http://www.flickr.com/photos/2768312...57622484352534

Right now my answer would be "not enough" (see the following photos if
you are interested in the process).

Can anybody suggest how such force could be measured?

Michael Koblic,
Campbell River, BC

Use a trailer tougue scale or a Snap on brake force gage. I happen to
have a snap-on brake force gauge set, they are only 1/2" thick and
made to go in a caliper in place of the brake pads to check caliper
pressure. Measure 0-5000 pounds.

http://www.sherline.com/lm.htm



Thank you and all the others for helpful information.

To answer some of the points that have been raised:

1) There is no question in my mind that two *heated* rollers would be
the way to go. I have looked at several solutions including laminators
but nothing fit the bill.

2) Up to now I have been heating the workpieces on the same 1/4"
aluminum plate and then going over them with a hard rubber roller.
Using a bathroom scale etc. I estimated the pressure developed this
way at 21 psi. The results are reasonable with temperatures in the
region of 160-170C but the bubbles are a problem as the pressure is
applied *after* heating. No amount of rolling will get rid of them
then. Often they are in a place where they can be re-touched but
sometimes they are not and the piece is scrap.

3) The reason I do it this way after five years of experimenting is
that I found the recommended method by Press-n-Peel quite useless. It
works on PCBs. On anything thicker - not so much. There are others who
have described successes with the Press-n-Peel iron-on method but the
big question is always consistency. I did look at the Hobby Lite press
from HIX but before I spent $325 I wanted to make sure that the press
is suitable for what I wanted it for. The company was not particularly
helpful or forthcoming with information.

4) I jury rigged a press using two clamps. There is no question that
on small pieces at least I develop higher pressures this way. This has
reduced the number and size of the bubbles. However, the rub is that
sometimes the pressure is too much and fine detail gets obliterated (I
did this today with a 7 in2 piece which I only took up to 130C - one
tiny bubble but I shall have to go over the details with a needle).

5) I knocked my version of the press together from whatever I had in
the house. Frankly I did not expected it to last very long. Using the
two clamps was a pain and sometimes there was a clearly discernible
pressure gradient across the piece. I was hoping that the central
screw will provide more even pressure distribution and better pressure
control. I was disappointed at its performance but in retrospect I
should have expected that as the whole area under pressure is 80 in2
thus to produce the same pressure as I do manually with the roller I
would have to develop over 1600 lbs force. Of course the idea was to
produce more.

6) The press reminded me today that the force developed is not
inconsequential as it blew one of its legs off. Still, I learned a
fair bit from the effort.

7) Thanks to the gentlemen who provided me with the formula. I found
it independently in the Machinery' Book late last night after I
posted. I should have found it much sooner if I fed "jackscrew" into
Google. Such is life :-). The figures I got was 125x multiplier
without friction. I tried to do the calculations with an assumed
coefficient of friction of 0.2 and got a multiplier of 31.5. Seems
kind of small. I haven't the foggiest what kind of force I put on the
end of the lever (3" long, BTW, the pitch is 0.151" and the pitch
diameter roughly 0.6").

8) Version 002 is in the works with many changes. I am still not sure
how to achieve a consistent force (it is much easier to manipulate the
temperature). I wonder if a torque wrench would be the answer if
somehow incorporated in the top lever.

Michael Koblic,
Campbell River, BC

On Jul 15, 10:01*pm, wrote:
The figures I got was 125x multiplier
without friction. I tried to do the calculations with an assumed
coefficient of friction of 0.2 and got a multiplier of 31.5. Seems
kind of small.
Michael Koblic,
Campbell River, BC

My father tells me that anything that will not run backwards has an
efficiency of less than 50%.
You cannot spin the shaft on a vise by prying the jaws apart, or it
would not stay clamped.

He once designed a vehicle with worm gear drive, that when the driver
took his foot off the gas, the vehicle skidded to a stop.

I'm kinda' late to this discussion, but I just had an idea for a device
that would excel in evenly-applied-force. That would be an air bag,
fitted in a box. A reverse of the vacuum air bag, which is limited to
15 psi. I don't know how much more that 21 psi you'd want, but a
bag-in-a-box could do much more than that.

For a bag, how about the bladder that's used in the storage tank for
well water, or in the expansion tank of hot water heating system. The
bladder itself is available as a replacement part.

Bob

On Sat, 16 Jul 2011 14:22:33 -0400, Bob Engelhardt
wrote:

I'm kinda' late to this discussion, but I just had an idea for a device
that would excel in evenly-applied-force. That would be an air bag,
fitted in a box. A reverse of the vacuum air bag, which is limited to
15 psi. I don't know how much more that 21 psi you'd want, but a
bag-in-a-box could do much more than that.

For a bag, how about the bladder that's used in the storage tank for
well water, or in the expansion tank of hot water heating system. The
bladder itself is available as a replacement part.


Nice idea. I like vacuum techniques. I use small vacuum bags to glue
these same finished workpieces to steel substrate rings. I found that
it worked better than clamps to get accurate positioning.

However, in this application it would have to be a bag that withstands
200C temperature.

Michael Koblic,
Campbell River, BC

On Jul 16, 1:01*am, wrote:
...
7) Thanks to the gentlemen who provided me with the formula. I found
it independently in the Machinery' Book late last night after I
posted. I should have found it much sooner if I fed "jackscrew" into
Google. Such is life :-). The figures I got was 125x multiplier
without friction. I tried to do the calculations with an assumed
coefficient of friction of 0.2 and got a multiplier of 31.5. Seems
kind of small. I haven't the foggiest what kind of force I put on the
end of the lever (3" long, BTW, the pitch is 0.151" and the pitch
diameter roughly 0.6").

Michael Koblic,

I think up to ~80% of the tightening force on a screw can be lost to
friction. Compare the torque to tighten with that to loosen.

jsw

On Sat, 16 Jul 2011 15:24:04 -0700 (PDT), Jim Wilkins
wrote:

On Jul 16, 1:01*am, wrote:
...
7) Thanks to the gentlemen who provided me with the formula. I found
it independently in the Machinery' Book late last night after I
posted. I should have found it much sooner if I fed "jackscrew" into
Google. Such is life :-). The figures I got was 125x multiplier
without friction. I tried to do the calculations with an assumed
coefficient of friction of 0.2 and got a multiplier of 31.5. Seems
kind of small. I haven't the foggiest what kind of force I put on the
end of the lever (3" long, BTW, the pitch is 0.151" and the pitch
diameter roughly 0.6").

Michael Koblic,

I think up to ~80% of the tightening force on a screw can be lost to
friction. Compare the torque to tighten with that to loosen.


I guess they are different :-) ??

I re-did the calculations using the 2 formulae in the Machinery's
Handbook: One for when the motion is in direction of the load
(assisting it) and the other in the opposite direction (opposing).

With the same data I got the force necessary to move a load of 1000 lb
to be 6.66 lb in the first case and 9.68 lb in the second case. This
time the multiplier is 103 and 150. GOK where that 31.5 came from. The
multiplier without friction is smack in the middle. That makes sense,
does it not?

Michael Koblic,
Campbell River, BC

On Jul 16, 11:18*pm, wrote:
...
I think up to ~80% of the tightening force on a screw can be lost to
friction. Compare the torque to tighten with that to loosen.

I guess they are different :-) ??
...
Michael Koblic,
Campbell River, BC-

Not by much sometimes. AFAIK the torque to loosen is mainly from
friction. In a simple system like sliding a load up a ramp the force
going up is the friction + the load, going down it's the friction -
the load.

My 10,000 # load cell is a pressure gage on a cylinder like this:
http://www.harborfreight.com/10-ton-...ram-95979.html

You might find a spring scale for weighing fish/deer/moose in a
sporting goods store.
http://www.harborfreight.com/hand-he...ale-97227.html

jsw

On Jul 16, 6:24*pm, Jim Wilkins wrote:

I think up to ~80% of the tightening force on a screw can be lost to
friction. Compare the torque to tighten with that to loosen.

jsw

This seems too obvious to mention, but I will anyway. The friction
depends on the lubrication. A ball bearing thrust bearing or a roller
bearing would also help.

You can find a lot of the type of information you want in web sites
that are about torquing bolts. Especially on the effects of
lubrication.

Perhaps you could incorporate Belleville washers to give you an idea
of the force applied. Belleville washers can be stacked to increase
the force necessary to flatten them.

Dan

On Jul 17, 7:53*am, " wrote:
...
This seems too obvious to mention, but I will anyway. *The friction
depends on the lubrication. *A ball bearing thrust bearing or a roller
bearing would also help....

Perhaps you could incorporate Belleville washers to give you an idea
of the force applied. *Belleville washers can be stacked to increase
the force necessary to flatten them.

* * * * * * * * * * * * * * * * * * * * * * * * * * Dan

IIRC experiments on torqueing aircraft fasteners showed a 2:1
variation in clamping force vs torque from surface roughness etc under
the best conditions. A skilled mechanic's judgement was as good as a
torque wrench. The best way to torque a fastener to a large fraction
of its yield point was to measure the change in length, though that's
not always possible. I saw that in print a long time ago.

The bottom line is you have no good way to predict what's happening
and should measure the forces.

jsw

On 2011-07-15, wrote:
How much force does a common shop vise develop in its jaws? Google was
pretty unhelpful on this - the only reference I could find suggested
up to 7000 lb. Another paper suggested that a minimum required
clamping force for machining is much less - up to 1000 N.

Well ... just as a point of information, I clamped a short
Enerpac hydraulic cylinder rated for 10 tons at 10,000 PSI in an old
Bridgeport milling vise. (I have Kurt vises which I prefer to use.)

I connected it to an electric hydraulic pump intended for
running large terminal crimpers.

That pump is designed to go up to 8400 PSI, and then release.

When I ran it on the Enerpac cylinder, I could *see* the frame
of the vise bowing -- perhaps the center lifted about 1/8" above the
line between the ends (this is without the vise being clamped to
anything -- just resting on wood). So -- at 8.400 tons (16,800 PSI) it
was bowing well beyond normal operation. so let's assume something much
more reasonable in such a heavy vise -- say about 1 ton or less. (And I
would have had to crank the handle a lot harder than I can physically
manage without a long cheater pipe to get to that, I believe.

As the vise returned to straight, it was within the elastic
range.

I am interested because I am trying to guess how much force I can
develop in this press:

http://www.flickr.com/photos/2768312...57622484352534

Right now my answer would be "not enough" (see the following photos if
you are interested in the process).

Can anybody suggest how such force could be measured?

The first question is -- what force are you trying to measure?
The easiest to measure is the force applied by the jackscrew to the
middle of the plate. There are load cells which display the force
applied as the frame deflects (and it is measured by a dial indicator).

However -- this is not measuring the force applied to the
transfer labels, which is what I think you want. Just knowing the force
applied by the rod is not enough, because both your backing plate and
your top pressure plate are far from rigid enough.

For the top plate, you need something like forged steel with a
platform in the middle (at a guess say 2-3" high) with ribs going out
from there to the corners, and angling down to the corners. This will
even out the force significantly.

However -- your aluminum plate on the bottom also bows, reducing
the pressure in the middle and concentrating a bit more of it out to the
edges.

What I would suggest is:

1) Triple the thickness of the bottom plate, and make it steel,
not aluminum.

2) Go up to at least one inch thickness on the wood top plate,
and ideally at least two.

3) Get a closed cell foam rubber to go between the wood pressure
plate and your transfers. This will crush, and even out the
force from center to edge significantly. At a guess, I would
suggest perhaps an inch thickness or more for the foam rubber.

With this, you probably won't need as much force as you were
applying in your tests.

Good Luck,
DoN.


--
Remove oil spill source from e-mail
Email: | Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---

On 17 Jul 2011 02:00:00 GMT, "DoN. Nichols"
wrote:

[...]

I am interested because I am trying to guess how much force I can
develop in this press:

http://www.flickr.com/photos/2768312...57622484352534

Right now my answer would be "not enough" (see the following photos if
you are interested in the process).

Can anybody suggest how such force could be measured?

The first question is -- what force are you trying to measure?
The easiest to measure is the force applied by the jackscrew to the
middle of the plate. There are load cells which display the force
applied as the frame deflects (and it is measured by a dial indicator).

This would do fine. All I am after is something repeatable. The actual
pressure applied across the workpiece can be calculated from the
surface area.

However -- this is not measuring the force applied to the
transfer labels, which is what I think you want. Just knowing the force
applied by the rod is not enough, because both your backing plate and
your top pressure plate are far from rigid enough.

For the top plate, you need something like forged steel with a
platform in the middle (at a guess say 2-3" high) with ribs going out
from there to the corners, and angling down to the corners. This will
even out the force significantly.

There is one commercially available like that. I nearly bought it but
it was 3x as expensive and quite big (the leg span was 18").

However -- your aluminum plate on the bottom also bows, reducing
the pressure in the middle and concentrating a bit more of it out to the
edges.

That I do not understand. The reason it bows in the middle is because
the pressure is applied *there*.

What I would suggest is:

1) Triple the thickness of the bottom plate, and make it steel,
not aluminum.

Agreed except for the steel heat conductivity.

2) Go up to at least one inch thickness on the wood top plate,
and ideally at least two.

Probably also wise.

3) Get a closed cell foam rubber to go between the wood pressure
plate and your transfers. This will crush, and even out the
force from center to edge significantly. At a guess, I would
suggest perhaps an inch thickness or more for the foam rubber.

That's why I use the silicon pad. I wonder if ordinary rubber would
withstand the temperatures. I know hockey puck does not!

With this, you probably won't need as much force as you were
applying in your tests.

I know I can develop forces well in excess of required for the small
pieces (up to 16 in2). I suspect I shall never need to apply the same
pressure over the whole of 80 in2. Right now I suspect consistency
will be a bigger issue.

Michael Koblic,
Campbell River, BC