Miks15
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| posted on 21/2/10 at 09:49 PM |
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constant force from actuators?
For a uni project i need to design a linear wear testing machine with a pin rubbing on a plate but there must be a constant force of 500N being
applied to the top of the pin.
Can linear actuators being controlled this precisely to hold a constant force?
It onl has to be designed and not made, so as long as its doable then ill use that.
Or any other suggestions of how to apply a constant force?
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Breaker
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| posted on 21/2/10 at 10:01 PM |
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Just put 500N of weight (f.e. 2 sierra diffs) on the pin ? 
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MikeRJ
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| posted on 21/2/10 at 10:17 PM |
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quote:
Originally posted by Breaker
Just put 500N of weight (f.e. 2 sierra diffs) on the pin ?
That would be by far the simplest method.
A long spring (or shorter spring applying force through a long lever) making the ratio of pin wear to spring movement very small would also hold an
approximately constant force.
You can get constant force springs (think tape measure, inertia reel seat belt mechanism), but I don't know if any would be large enough to
provide 500N.
If you want to do it electronically things get a lot more complex, expensive and unreliable.
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brianthemagical
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| posted on 21/2/10 at 10:38 PM |
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Hydraulics?
Just need some kind of simple pump controled by a pressure switch and you're done. depends how accutrate/toleranced it needs to be.
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matt_gsxr
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| posted on 21/2/10 at 10:40 PM |
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Pneumatic cylinder. Pressure is easily regulated and monitored.
But why not a 50kg weight.
I guess I don't understand the problem well enough.
Matt
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Miks15
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| posted on 21/2/10 at 10:45 PM |
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wel i guess a 50kg weight would be the simplest option, one we did think about and might still use, just wondering if its possible to do it with
either electronic/pneumatic or hydralics actuators
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indykid
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| posted on 21/2/10 at 10:46 PM |
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if you don't want to add the 51kg to the top of the pin, assuming it would be reciprocating, 2 identical hydraulic cylinders one rigidly fixed
and loaded with 51kg, then a flexi to the other cylinder on the rig.
minimal reciprocating mass and constant force application.
tom
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indykid
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| posted on 21/2/10 at 10:49 PM |
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quote:
Originally posted by Miks15
just wondering if its possible to do it with either electronic/pneumatic or hydralics actuators
it is possible, but you would need a load cell on the test piece and a closed loop control system
adding the mass one way or another is by far the most robust method. always make test rigs as robust as you feasibly can!
there's no point getting to the last 1/5th of the test and finding it's gone out of tolerance.
tom
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MakeEverything
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| posted on 21/2/10 at 11:05 PM |
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Is the pin reciprocating, or is the wear plate turning / reciprocating?
Add the weight to the pin using a series of CF springs. This will be the most compact method and can be used with either the wear plate or the pin
moving.
Kindest Regards,
Richard.
...You can make it foolProof, but youll never make it Idiot Proof!...
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Miks15
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| posted on 21/2/10 at 11:29 PM |
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the plate will be reciprocating, with the pin placed on top
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watsonpj
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| posted on 22/2/10 at 01:21 AM |
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The answer to whether you could do it with pneumatics,hydralics or electrics is almost certainly yes but the best solution starts with the
specification.You say you need 500N but is that 500+/- 1 or 50. Lecturers generally set very open ending questions but when you approach these kinds
of problems you need to clarify the spec to understand the requirements and therefore what you can get away with. A weight my be adequete but if the
plate moves quickly the additional accel decel may well push it out of range.
I'd have a think (brainstorm if a group) on what you could use and also go back on the spec and question it and then you can start working your
list.
good luck
Pete
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Miks15
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| posted on 22/2/10 at 06:46 AM |
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well like you said the spec is very open ended, just saying a force of 500n needs to be placed on top.
We dont need to design anyof the electrical compnents so we can just say computer controlled to maintain a 500n force on top.
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indykid
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| posted on 22/2/10 at 09:53 AM |
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it wouldn't be much more effort to include some more detail and show you actually understand the problem. that's where the marks are
are you in first year? if so, i suppose you can go for easy street....if not though, get it in there and get yourself some marks. better first year
grades help you get better placements if you're on a sandwich degree too
tom
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britishtrident
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| posted on 22/2/10 at 10:51 AM |
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Remember the KIS principle ---- don't get involved in electronics or hydraulic or pneumatic regulators simply use a weight. If don't want
sit the weight directly on top of the pin apply it remotely by levers or hydraulics - as used in a Budenberg Dead Weight Tester used for calibrating
pressure gauges.
[I] “ What use our work, Bennet, if we cannot care for those we love? .”
― From BBC TV/Amazon's Ripper Street.
[/I]
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Miks15
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| posted on 22/2/10 at 01:47 PM |
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I think were just going to go for a weight on an arm, to reduce the actual weight we have to have in there, cheers guys
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hughpinder
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| posted on 22/2/10 at 01:53 PM |
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Just some things to think about:
Is the reciprocating plate flat and even thickness - If the height of the top surface varies a lot (e.g. curved) and you move it quickly, you will
have inertia of the weights to consider.
Is the plate strong - how is it supported - a 50kg point load would produce a fair flex even in a fairly strong plate, obviously depending on length
and shape of the sample.
You may be able to devise something with two arms pivoted at the end,with one each side of the plate, a wide wheel on the lower one and the wear pin
on the other arm. Pull the arms together with a pneumatic/hydraulic actuator (allowing for leverage), and top up the pressure to keep the force the
same as the pin wears out (or if the plate is not even thickness).
Have you ever been to Ikea? They have various wear testing machines on display that may give you some ideas (or they did 5 years ago when I was there
last).
Just some things to think about,
Regards
Hugh
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iank
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| posted on 22/2/10 at 02:58 PM |
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Depends on what they want is the real answer, suggesting a fixed mass on top of the pin is fine but might not get many marks if it's at the end
of a hydraulic control system module 
Remember the critical evaluation of your solution is the thing that gets the marks (things like inertia of a 50kg weight will need to be worked out if
you're presenting that as a solution).
It's not hard to do with hydraulics if I remember stuff I did 20 years ago. I think you need a load sensor between the pin and the cylinder and
some kind of simple control algorithm. Would probably be best to have a Kalman filter in there if you're happy to talk about them.
The advantage of a computer controlled system is it's a lot more compact at the test head end and you can change the force applied on the fly if
it's necessary.
--
Never argue with an idiot. They drag you down to their level, then beat you with experience.
Anonymous
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britishtrident
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| posted on 22/2/10 at 03:47 PM |
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Most complex solution is never the best.
Simplest solution is most often the best.
[I] “ What use our work, Bennet, if we cannot care for those we love? .”
― From BBC TV/Amazon's Ripper Street.
[/I]
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MikeRJ
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| posted on 22/2/10 at 08:22 PM |
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quote:
Originally posted by iank
Remember the critical evaluation of your solution is the thing that gets the marks (things like inertia of a 50kg weight will need to be worked out if
you're presenting that as a solution).
Sounded to me like the pin was stationary and the plate was being moved underneath it?
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iank
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| posted on 22/2/10 at 09:31 PM |
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quote:
Originally posted by MikeRJ
quote:
Originally posted by iank
Remember the critical evaluation of your solution is the thing that gets the marks (things like inertia of a 50kg weight will need to be worked out if
you're presenting that as a solution).
Sounded to me like the pin was stationary and the plate was being moved underneath it?
Yes, but if the plate has anything that looks like a wear ridge the pin will bounce up (or down) it and the force may drop below and then increase
above 500N (or vice versa) before returning, in a real system it will then oscillate. Maybe a simple damper would be wise if that approach is being
taken? Though no system isn't ever going to be completely constant in all circumstances so you're going to need to define an acceptable
fluctuation in force that is considered 'constant'.
While the big weight may be the best solution I have a doubt that it's what a degree level course prof has in mind as the solution, unless
it's a trick question to weed out the people who over design of course.
[Edited on 22/2/10 by iank]
[Edited on 22/2/10 by iank]
--
Never argue with an idiot. They drag you down to their level, then beat you with experience.
Anonymous
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indykid
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| posted on 22/2/10 at 11:37 PM |
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quote:
Originally posted by iank
While the big weight may be the best solution I have a doubt that it's what a degree level course prof has in mind as the solution, unless
it's a trick question to weed out the people who over design of course.
the marks are most probably in choosing the best solution, whatever that may be, then justifying it and discussing the limitations.
a pure simple solution is something many of the lads on my course would be totally unable to conceive.
just because its uni, it doesn't mean it needs to be complicated. it may seem like common sense, but we have it. some people need a shove
tom
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iank
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| posted on 23/2/10 at 07:49 AM |
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quote:
Originally posted by indykid
quote:
Originally posted by iank
While the big weight may be the best solution I have a doubt that it's what a degree level course prof has in mind as the solution, unless
it's a trick question to weed out the people who over design of course.
the marks are most probably in choosing the best solution, whatever that may be, then justifying it and discussing the limitations.
a pure simple solution is something many of the lads on my course would be totally unable to conceive.
just because its uni, it doesn't mean it needs to be complicated. it may seem like common sense, but we have it. some people need a shove
tom
Seems's we're on the same page with where the marks are earned
quote:
Originally posted by iank
Remember the critical evaluation of your solution is the thing that gets the marks.
Paraphrasing Einstein "Everything should be made as simple as possible, but no simpler." The lump'o'lead solution does have
some problems that can't be ignored depending on the original phrasing of the problem and the prejudices of the person marking it.
p.s. I've been working for 20 years in the software industry and your second sentence made me smile and wince at the same time!
--
Never argue with an idiot. They drag you down to their level, then beat you with experience.
Anonymous
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