Help , can someone provide me with some equations !

[Harold frankensteinParticipant @haroldfrankenstein96209]

anyone please knows how to calculate the load that this worm gear is about to get from the mass of both the table and the piece , knowing that the operator has to apply a force enough to turn it too , these are the settings of the gear :

[Harold frankensteinParticipant @haroldfrankenstein96209]

[DC31kParticipant @dc31k]

You might have to expand on the context of your question. Forgive me if I am wrong, but it initially smells like you are asking us to do your homework for you.

One thing is for sure, the load on the worm gear will vary a lot. When the table is horizontal, it will see very little load. When the table is at 40 degrees and you are trying to move it back to horizontal, it will see a considerable load.

[JasonBModerator @jasonb]

The actual shape of your 200kg load will also play a large part in what is required to move the table as the further from the pivot point the more effort will be needed.

I would have thought the formula needs some trigonometry to take in the angular position of the combined load and table probably best to do it in two one calculation for the table and a second one based on the centre of gravity of the workpiece then add the two together.

[Martin ConnellyParticipant @martinconnelly55370]

The mechanical advantage of the worm is the circumference of the operating handle rotation divided by the pitch of the worm. So one rev of the handle will be 628mm and the worm pitch is 15.7mm giving a mechanical advantage of 40:1. Seems a very coincidental round figure, sounds like a homework question.

The next part is the leverage of the rotation about the centre of rotation of the mass. The worm force is acting 200mm away from this at the pitch circle radius of the gear. The centre of mass of the load and table is about double this from the centre of rotation so mechanical dis-advantage is about 1:2

So far we have a mechanical advantage of about 20:1

Then there is the calculation of the force needed to move the mass. This is going to be given by mg (mass x gravity) but is depending on the position of the mass relative to the centre of rotation. So the working load mg needs to be multiplied by sin( angle from vertical) of the mass. So in the vertical position there is next to no force (barring mechanical losses due to friction for example) needed to rotate the load. In the horizontal position you will need mg x sin(90° )/20 newtons to push the handle around.

So at the worst position you will need about 286 x 10 x 1 / 20 = 143 (+friction) Newtons to push the handle around, a torque of 14.3Nm. With added friction maybe 15Nm

So say 150N to push the handle round which is about the same as lifting a 15kg weight.

I have a suspicion there may need to be some calculation added because of the diameter of the worm. But this may be cyclic depending on the position of the handle so will not be constant.

Some rounding has been used.

Of course I could be completely wrong but this is what I would have used at work if I was doing a feasibility calculation on something like this. I've still got the drawings of something for rotating a 850kg mass using a similar system but the out of balance weight was much lower as it was designed to rotate about the centre of mass.

Martin C

Smiley strikes again

Edited By Martin Connelly on 28/05/2021 08:47:56

[john carruthersParticipant @johncarruthers46255]

It doesn't apply with your rubric, but a counter weight might simplify the working out?
Like a telescope drive.

[pgk pgkParticipant @pgkpgk17461]

Asa smple-mnded soul i'd use a spreadsheet to chart a selection of positions and then see what is, is not linear noting that the lump on the table appears not to be bolted down so you ahve to consider coefficient of static v sliding friction too as an end point.

pgk

[Neil AParticipant @neila]

I think you need to be very careful when using a worm and worm wheel with a heavy out of balance load.

I have seen theoretically non-reversible worm and worm wheel arrangements run away on an large engine turnover frame when the load had not been balanced out sufficiently with counter weights. Quite scary moments for the people using it.

Neil

[Neil AParticipant @neila]

Duplicate post, sorry.

Neil

Edited By Neil A on 30/05/2021 11:51:16

[Nigel Graham 2Participant @nigelgraham2]

That's not a model-engineering hobby beginner's question, is it! Or is it?

A degree-course exercise perhaps. Or?..

The illustration is clearly diagrammatic only, possibly not part of a real design. The data may have been selected to simplify the arithmetic, but the photographed document is not a text-book page. The illustration may be from a book – but not necessarily on this topic. A CAD exercise maybe.

Where are the equations and value abbreviations? I suggest in the text-books used by the students for whom this problem is or was originally, intended.

Why the steel specification? Irrelevant to Harold's immediate question, the full exercise may also examine the individual components' strengths.

'

What is C45 steel anyway? This shows very poor practice by the question not using the standard format. C45 appears a brand-name used by the British company, Murray Steel Products, for a medium-carbon alloy, though no doubt with direct BS-EN / ISO equivalents.

'

I reckon one of two possibilities I call upon Harold to reveal.

– Is your question "homework" from a larger exercise assembled by your tutor from his course's set text-books and a trade catalogue? In which case surely you have the books and steel specifications?

OR …

– are you trying to design and make an adjustable angle-plate with a quarter-tonne SWL, and itself possibly at least as heavy, from very incomplete sources of background theory?

If the latter I suggest with great respect you are dropping yourself in at the deep end. You would be better off either buying commercially available equivalents (at great cost) or finding a proven design for your particular applications and facilities. Are you really planning to machine 200kg components to fine limits? That is some workshop you own! Or are you using a college-tutor's question as a basis for something like a rotating assembly-bench; for which various designs have appeared occasionally in ME?

[JasonBModerator @jasonb]

Posted by Nigel Graham 2 on 31/05/2021 13:01:46:

What is C45 steel anyway? This shows very poor practice by the question not using the standard format. C45 appears a brand-name used by the British company, Murray Steel Products, for a medium-carbon alloy, though no doubt with direct BS-EN / ISO equivalents.

Its an EN specification so no need for equivalents and quite commonly used . EN 10083 if you want a read page 16 lists all the various grades

[SillyOldDufferModerator @sillyoldduffer]

Posted by Nigel Graham 2 on 31/05/2021 13:01:46:

That's not a model-engineering hobby beginner's question, is it! Or is it?

…

Personally, I don't care! I see anything of Engineering interest as fair game. and don't have strong views about the words 'beginner', 'hobby', 'engineering' or 'model'. Even if it is a homework question, I've found the answers useful.

C45 is educational too. Playing with FreeCAD's Finite Element Method workbench as a hobbyist beginner working on a model last week, I found most of steel material cards are for EN / ISO specs, which I'm not familiar with, and had to swot up on. Fortunately it's on the internet, though it took me a while to find EN3A is C15.

Relevant I suggest to model engineers because I don't know how long the old British Specs are going to last. They're not a problem for small users buying in the UK yet, but the rest of the world is drifting away from them. Older UK steel specs aren't supported by FreeCAD and I suspect other software ignores them too. When all us old chaps have been replaced by youngsters brought up on CAD, I suspect the hobby will be baffled by the likes of EN1A, just as most of us have forgotten 'Wednesbury Skelp' and 'Sham damn' Iron!

Dave

[Jeff DaymanParticipant @jeffdayman43397]

I hope the shown gizmo never gets constructed as drawn or there may be a bad shop accident. Certainly VERY expensive to custom make as drawn. In industry this sort of thing would likely be designed around a commercially made high ratio reducer worm gearbox, only the workholder table would be custom made, likely, and mounted securely to the reducer's output shaft , trunnion plate, or flange.

[Martin ConnellyParticipant @martinconnelly55370]

Jeff, this is a minor sized gizmo. Look up welding positioners and rotators and you will find plenty using similar principles for much larger masses. The larger ones do not use hand cranks though.

Martin C

[Nigel Graham 2Participant @nigelgraham2]

Jason –

Thank you for that information about specification.

As I say have never come across such a format, but an search yielded a steel company'e site in whichy the numner "C15" looked for all the worls like their own code. No mention of standards.

I used to work (as a materials store-keeper) to a MacReady's wall-chart listing all manner of grades by EN-number. I looked at an American code index, I searched my text-books for EN, ASME and ISO codes, and they all list no steels by just 3 characters.

So is this a new system, or abbreviation of an existing one?

Be that as it may I am puzzled why Harold is trying to work to a set of values given algebraic names obviously related to formulae, but does not have those formulae.

Jeff –

The drawing does not look as if intended as the full design, and in fact appears truncated, but to examine the basic form and principles from which to determine the calculations and physical details.

[Jouke van der VeenParticipant @joukevanderveen72935]

What about Topic Starter Frankenstein. Did he disappear?

[Author]

Posts