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How shallow an angle of V would still align with a round metal bar?
3

How shallow an angle of V would still align with a round metal bar?

How shallow an angle of V would still align with a round metal bar?

(OP)


Hello

I am new. I have simple a mechanical design problem:

1. I have a cylinder - i.e. a round bar - that is 25mm in diameter and made of polished metal (chrome).

2. I have an inverted V shaped guide made from smooth plastic (probably polyethyline PE or similar. This inverted V will be fairly short - only 10mm (1cm) deep.

Now, the plastic inverted V shaped guide will be resting (unsupported) on top of the round bar, with a light force equivalent to say 200 grams, and it will be able to twist along a vertical axis.

My question is this:
How shallow can I allow the angle of inverted 'V' to be and still expect the V to rotate so as to align with the direction of the metal bar? i.e. to point in the same direction. (The alignment of the two objects only needs to be within a few degrees)

To explain:
If the V is extremely steep say 45 degrees - or even 90 degrees - then it would be reasonable to expect the V to align EXACTLY with the bar. But if the V is very shallow and starts to approach 180 degrees, then one would not expect the V to rotate at all.

Are there any rules of thumb or guestimates that I could use?

J

P.S. Please excuse the very crude nature of the attached diagram, but I hope it brings the problem to life...



RE: How shallow an angle of V would still align with a round metal bar?

I do not know of any rule of thumb. I would expect the coefficient of friction would determine when the V will no longer, reliably align with the cylinder. The aligning forces, given your downward force, would no longer exceed the friction forces and there would not be enough force to cause the V to rotate to align.

Ted

RE: How shallow an angle of V would still align with a round metal bar?

What is preventing the round rods from rotating? THAT will determine how force you need to keep the angle iron and the bar aligned against the resisting torque.

For example: Short, round bar is on a sheet of smooth ice, at melting point. Almost no resisting force at all.

Long round bar is welded (glued) to a piece of steel plate. Near-infinite resistance to torque.

RE: How shallow an angle of V would still align with a round metal bar?

(OP)
Apologies it seems I wasn't clear enough.

> I would expect the coefficient of friction would determine when the V will no longer, reliably align with the cylinder.
With all due respect, this is not telling me anything new! What I need is to make a reasonable 'guestimate' for my first prototype.
i.e. Yes, the coefficient of friction is important - clearly. Fwiw, in this case I said it was PE - polyethylene - which as we all know has a very low coefficient of friction. However all the variables that I have supplied are important, including the depth of the V. If the v was very long, and if the V was a tight angle, then would become very easy. etc.

However in this case, the dept of the V is fixed at 1cm. The only real variable I have to play with is the angle of the V. And in order to save vertical space I need the V to be wide as possible...

To get clear, what I need help with is having a reasonable stab at what the design should be. Otherwise I will simply have to build endless, endless prototypes. And surely the whole point of the discipline of engineering is NOT to have to build endless prototypes in order to design things!

> What is preventing the round rods from rotating?
No, the rods are fixed in place and can not rotate at all. The V shaped plastic guide can rotate only a vertical axis. They can't rotate around a horizontal axis either. (To get clear, the reason for this is that the plastic V forms what is in effect the top section of a "hook" which goes over the metal bar. And the hook is kept in place by gravity. What I need to achieve is to find a way to make sure that the plastic hook ends up at EXACTLY (or almost exactly) 90 degrees to the the metal rod.

The only things that are stopping the V from rotating are:
a) Friction
b) If the plastic V rotates far enough around a vertical axis, eventually sides of the V will of course start to cross over the horizontal metal bar in a significant manner, and at this point to rotate any further would require the plastic V to start to be raised. Due to the weight at the other end of the hook this would require work to be done against gravity.

This is for an indoors consumer product. So there sill be little or now wind. Humans may occasionally bush very lightly against the plastic.


I think we are over-complicating this. Just want a V to be aligned along the side of a cylinder. And I want to know should it be: 5, 10, 15, 20, 30, 40 or 50 degrees!

J

RE: How shallow an angle of V would still align with a round metal bar?

The rod is smooth, right?

Is the V on a shaft or something like it so it will drop smoothly down onto the middle of the rod every time?

RE: How shallow an angle of V would still align with a round metal bar?

(OP)
The plastic V guide is smooth - probably waxy poly ethylene.

The metal bar will be a fixed object - normally polished metal. (Although it may possibly be smooth painted/varnished wood.)

The V shaped plastic guide is the top section of a hook design and is held in place purely by gravity.

RE: How shallow an angle of V would still align with a round metal bar?

Make the included angle 120 degrees. The ramp out force will be .58 x the weight.

Ted

RE: How shallow an angle of V would still align with a round metal bar?

(OP)
> Make the included angle 120 degrees.
Sounds reasonable...
I'm slightly confused about the rationale for 120 though.
How far will the vertical axis rotation need to proceed in order to get the your ramp out force?

RE: How shallow an angle of V would still align with a round metal bar?

Rotation beginning from being seated on the cylindrical surface.

Ted

RE: How shallow an angle of V would still align with a round metal bar?

If there is any magic number it would be a 90 degree included angle (45 degree ramp) At this point you have a balance between vertical and horizontal ramps. Sort of like a cannon that that get maximum range at a 45 degree angle.

RE: How shallow an angle of V would still align with a round metal bar?

(OP)
I agree that 45 degrees would work. However it does seems unnecessarily extreme. Like I say, I am after the widest angle that will still be likely to work.

I am a humble novice, but it's starting to sound like there are no rules of thumb or sensible guesses and that the engineering is too complicated to work out! :^(

RE: How shallow an angle of V would still align with a round metal bar?

You should figure out the relationship between V-angle and coefficient of friction that predicts sliding (if that's what you're interested in) or such a relationship that predicts no motion. Find some typical values for the materials and surface conditions you intend to use, then plug them in the relationship.

Getting values for dynamic and static coefficients of friction isn't always easy. I suspect your conditions will not be standard. The values you get will depend on surface condition in the contact areas.

This doesn't look like a setup anyone has equations for in their back pocket. I certainly don't. You may have to do that part on your own.

You will need to consider that the contact angle between the bar and angled plastic will change as the yaw angle (rotation) changes.

Once you get results from the above, build some and test them. (You might consider just skipping to this step anyway.)

I hope some of that is helpful. Good luck

RE: How shallow an angle of V would still align with a round metal bar?

It is evident to me that the contact pressure and hence contact area will be strong functions of the misalignment angle. Given the probable effect on net friction (assuming finite hardness and non-constant coefficient of friction with pressure), you will probably need a good friction model for your two materials in contact with each other, for starters.
You might try looking for the other constraints on your design, to give you an end point for the steepest angle that will perform as required, in order to narrow the design space. For instance, what is the maximum angular misalignment that is anticipated as a starting condition? Obviously, if there is an upper limit on the hieght of the V, a very steep angle will not be able to latch on in the first place, beyond a certain amount of misalignment, while a shallow angle approaching 180 deg. will be able to latch on, just barely, at an angle where the width of the V times the cosine of the misalignment angle is equal to 12.5mm. Also obviously, "just barely" is not a good criterion, so you'll need to determine a reasonable margin for latching on reliably. This is not the answer, just some points you might consider along the way to finding it.

"Schiefgehen wird, was schiefgehen kann" - das Murphygesetz

RE: How shallow an angle of V would still align with a round metal bar?

If there is no friction opposing rotation of the guide, and for small angles of mis-alignment, the problem reverts to a simple "mass holding on a slope".
So mu = tan(slope) and included angle (your problem), I = 180 - 2 x slope. (I = included angle)
So I = 180 - 2 x tan^-1(mu)
eg
mu = 0.1 gives I = 168.6 deg
mu = 0.5 gives I = 126.9 deg
mu = 1.0 gives I = 90 deg

je suis charlie

RE: How shallow an angle of V would still align with a round metal bar?

Ship69, if there is a chance that the load could swing side to side, you may not want the shallowest angle. If too shallow, the load may not recenter itself or may fall off.

Ted

RE: How shallow an angle of V would still align with a round metal bar?

(OP)
> Ship69, if there is a chance that the load could swing side to side, you may not want the
> shallowest angle. If too shallow, the load may not recenter itself or may fall off.

No - not really, its movement will be heavily dampened (by air resistance), and so multiple swings back and forth will be unlikely.
On the other hand, it be put into place by hand and for this reason will be extremely likely to be put on somewhat squint (or slightly rotated incorrectly in all possible axes!). For this reason it needs to 'want' (via gravity) to align itself correctly at 90 degrees to bar even if placed incorrectly.

To get clear the plastic V is part of large hook with 90+% of the weight hanging below the horizontal metal bar, so there is close to zero chance of the hook falling off the metal bar.




RE: How shallow an angle of V would still align with a round metal bar?

(OP)
robyengIT - I had a quick look at your equations and they seem to be more about pipes joining each other at strange angles. Whereas whatever it is that I need friction will clearly be a core component in the calculations.


Unless someone can tell me otherwise, I fear that running what will probably be a large number of prototypes will be a lot quicker then trying to calculate the mechanics! :^(

OP

RE: How shallow an angle of V would still align with a round metal bar?

ship69, responders have given you angles and angles referenced to friction coefficients. What exactly are you looking for?

Ted

RE: How shallow an angle of V would still align with a round metal bar?

Looking at tables of friction coefficients, I see .20 as worst case. Therefore, minimum angle for impending slide is arctan(.2) = 11.3 degrees. The maximum included angle would be 180 - 22.6 = 157.4 degrees. The included angle must be less than 157.4 degrees in order for the V to slide to align with the cylinder.

Ted

RE: How shallow an angle of V would still align with a round metal bar?

(OP)
OK let me try and get my head around this.

1. Shallow 'compass bearing' angles
In the starting condition, if the top of the hook - i.e. the plastic (inverted) V - has been rotated horizontally to be aligned at a pretty extreme compass bearing (e.g. say 20 degrees) away from that of horizontal bar - then, given a reasonably steep V and a fairly low amount of friction between the plastic V against the metal bar then my assumption is that the V will definitely rotate towards a better alignment with the bar.

However the problem starts when the compass bearing between the V and the bar is more shallow, being as the corrective angles will be much more shallow.

2. Tighter alignment needed
I've run my geometry again and the bad news is that I need much tighter alignment than I had previously thought. Ideally as little as +/- 2 degrees would start to become a problem. So I am really looking the plastic V to come to rest at +/- 1.5 degree from the compass bearing of the bar.


3. Hydtools's calculations
I'm not quite sure what went into the 157.4 degrees calculation (it's is a pretty wide looking V either way). Is it the steepness of slop at which a down which an object (in simple terms think 'toboggan'!) would start to slide?

If so maybe we could find a way to use that. However the complication is that we need to get this angle calculated correctly - and this is NOT straightforward. Not least because the closer to true alignment of metal bar and V-shaped plastic, the smaller this angle will be. (This will be due to the fact that the plastic can only rotate around a vertical axis due to the weight at the bottom of the hook [although see Real World Experiments below, talking about swing.)

4. A possible simplification
One thing that might help is if we were to assume that the thickness/depth of the V was extremely narrow (approaching infinity - but still deep enough to have significant friction). Any increase in depth (Z direction) will help correct, so if something can work at a very thin (e.g. 1mm) Z depth, then it should work with my actual model.

5. Real world experiments ==> swinging
I have done some simple tests with a crude prototype and during the time when the weighted hook (containing the plastic V) has first been put on the metal bar, depending on exactly how crudely it has been hooked onto the metal bar, despite the fact that there is quite a lot of dampening by air, it will swing always back and forth a number of times. With fairly smooth surfaces this can easy be as many as 12 swings in the Y direction (see diagram). The air dampening in the Z direction is greater but that can swing a number of times too.

What I have only now released is that whilst any of this swinging is still happening, because things are already moving small corrections to the alignment become possible that would not have happened during static conditions.

[In hind-sight this is quite obvious - it's just that I hadn't realized how many swings would happen when the hook is weighted.


Here is a slightly improved diagram:



With thanks

OP

RE: How shallow an angle of V would still align with a round metal bar?

Extremely narrow z, approaching zero, not infinity but knife edge, there will be infinite z swinging neglecting air damping friction. Increasing z dimension will increase resistance to z swinging.
Decreasing included angle will increase aligning force. X swinging will be damped by sliding friction.

What are your post-hanging motion requirements? Something more than just aligning the V to the bar?

Observe clothes hanger motion for clues.

Ted

RE: How shallow an angle of V would still align with a round metal bar?

(OP)
> What are your post-hanging motion requirements?
> Something more than just aligning the V to the bar?
If the hook gets brushed lightly e.g. by hand, it would need to swing about a bit but again come to rest hanging (to within c.1.5 degrees) parallel to the horizontal bar.


> Observe clothes hanger motion for clues.
Yes - good call. The behaviour of the device will be pretty similar to that of a cloths hanger - with clothes on it.


P.S. CORRECTION: The diameter of the metal bar is 25mm (not 25cm) - my apologies.
I have corrected that previous diagram.


RE: How shallow an angle of V would still align with a round metal bar?

Ship69:
Wow! 25cm changes to 22mm, not much difference there. That certainly changes the proportions of the sketch and some of the considerations of the problem. The last few posts are really starting to get to the root of the problem. It seem to me that an improvement in the design would be, instead of the Vee shape at the apex, could you form that plastic top piece to match the o.d. of the rod. Then, if the included angle of actual contact was 10 or 15̊ on either side of the top of the rod (from 11 to 1 o’clock), it would have some real holding power and centering ability in each direction. RE: the coat hanger analogy, note also that while the head of the hanger doesn’t normally contact the rod except over a short length at the top of the rod, it does circle the rod from 9 o’clock to 4 o’clock, in terms of preventing falling off the rod. This would also allow you to significantly reduce the size of the cranks (lateral length dimensions, x direction) in the vert. portion of the hanger. And they could be nicer curved shapes (more generous curved shaped), rather than the 90̊, tight bends, which are very high stress regions. You would have better centering ability of the load, if the vert. length of the hanger was longer below the rod, if this is possible. What are you actually hanging from these hangers and how are they used. I find it hard to imagine that they might not see greater lateral loads than you suggest. How are they installed and removed from the rod? What is the area of the hanging article, as a sail to a slight wind? Can they be bumped?

RE: How shallow an angle of V would still align with a round metal bar?

(OP)
dhengr:
Humble apologies again for my order of magnitude typo.

Exact diameter is not under my control. If the diameter of the horizontal bar was known like would be easier(!)

However in practice, depending on exactly where the the product that I am designing will be used, it will need to hook onto a fairly wide range of sizes of horizontal bar. This could be anywhere between 45 and 10mm, although more normally the diameter will be somewhere between 15mm and 30mm.

Btw, please ignore other issues such as aesthetics and stress levels at angled joints. (I have already dramatically simplified the over-all actual design in order to focus the issues I have raised. Moreover I am constrained by IP issues so can't disclose the actual design here.)

The product is for indoor use. Yes, lateral loads may be temporarily be larger that I suggest, in which case too bad - the product will move out of position. However the whole point of this discussion is to design something with a sporting chance of getting back into the 'correct' position when it stops swinging. :)





RE: How shallow an angle of V would still align with a round metal bar?

The gizmo is not so complicated that it couldn't be prototyped. There is something to be said for walking out to the shop and getting somebody to make you one, or get one 3-d printed, or cobble one together in your garage.

RE: How shallow an angle of V would still align with a round metal bar?

It might be better to make two separate Vs, each of short length, as a continuous V of similar length might not be straight and can rock. The discontinuous V will have 2 pairs of contact points, while the continuous one will ideally form two lines of contact. Any irregularity along either of the lines will reduce the V to two points of contact, spoiling the control.

RE: How shallow an angle of V would still align with a round metal bar?

(OP)

Quote (3DDave)

It might be better to make two separate Vs, each of short length, as a continuous V of similar length might not be straight and can rock. The discontinuous V will have 2 pairs of contact points, while the continuous one will ideally form two lines of contact. Any irregularity along either of the lines will reduce the V to two points of contact, spoiling the control.
This sounds intriguing. I don't quite understand. Do you mean one V inside the other (i.e. of shorter length) or one V behind the other? (i.e. further away from the viewer in my above diagram)

OP

RE: How shallow an angle of V would still align with a round metal bar?

He is saying to cut out the middle portion of your vee because it does nothing, unless it is not the correct shape.

RE: How shallow an angle of V would still align with a round metal bar?

Take that idea further. Three legs, two on one side and one centered on the opposite side. Three-point contact is always stable.

Ted

RE: How shallow an angle of V would still align with a round metal bar?

Let's get the geometry sorted out first, and see where this leads us.  Assume (as everyone has, perhaps heroically, been assuming) that both parts are infinitely rigid.  Consider the geometry when the vee (whose included half-angle is alpha and whose "axial" length is L) is rotated by an angle beta relative to the bar (whose radius is R).

See my attached diagram for this situation.  (Unfortunately I do not know how to include the diagram in the body of my post, for which my apologies.)  It is simple trigonometry to calculate the height of the vee's ridge above the bar's centroidal axis.  This height (H) is the sum of the two components H1 and H2 shown on the diagram.

Now, for the moment, assume that everything is friction-free.  Under the action of gravity on its self-weight (W), the vee will want to slip back to its beta=zero position.  To prevent this from happening we need to apply a torque to the vee, a torque Tg about the vertical axis.  Using energy-conservation or virtual-work principles it is easy to show that the magnitude of this torque needs to be
Tg = W * d/db(H)
where d/db(H) is the first derivative of H with respect to beta.

I believe, but cannot readily verify mathematically, that this torque is smallest when beta is vanishingly close to zero.  Under these circumstances d/db(H) can be shown* to be
d/db(H) = L/(2*tan(alpha))
and so
Tg = WL/(2*tan(alpha))
(*Being lazy, I did the calculus using the venerable Derive computer algebra program.)

Another result that comes from simple statics is that vertical equilibrium requires that the (radially directed) normal contact force between the vee and the bar be given by
N = W / (2*sin(alpha*cos(beta)))
at each of the two contact points.

Now think about friction.  The key question is whether the friction is capable of generating a resisting torque that is equal to the gravity-induced restoring torque Tg (perhaps throwing in a factor of safety for good measure).  The maximum possible friction force at each of the contact points is u*N where u is the coefficient of friction and N is as given above.  This will be oriented directly opposite to the direction of the relative movement, but the direction of relative movement is not easily calculated.  Assume, probably slightly conservatively, that the direction is the one that leads to the maximum resisting torque.  The frictional resisting torque will then be given by
Tf = (u*N) * (distance between contact points)
   = (u*N) * ( L/cos(beta) + 2*R*tan(beta)*cos(alpha*cos(beta)) )
For our case of interest, the case where beta is vanishingly small, this simplifies to
Tf = uWL/(2*sin(alpha))

If the calculated value of Tg exceeds that of Tf by a sufficient margin, then we can be confident that the vee will correctly self-orient itself on the bar.

RE: How shallow an angle of V would still align with a round metal bar?

(OP)
Denial's image:

Fwiw, I couldn't get the PDF to work and so converted to GIF format.

I shall now be away - back in a few days.

OP

RE: How shallow an angle of V would still align with a round metal bar?

There is an error in my diagram.  The hand-written expression for H1 should be
L*sin(beta) / (2*tan(alpha*cos(beta)))

This problem applies only to the diagram, and results from a careless transcription error when I created the higher quality diagram from the crude scratchings I had used when doing my algebraic manipulations.  The algebra in the text of my above post is unaffected by this error.

RE: How shallow an angle of V would still align with a round metal bar?

One thing that needs to be answered is whether the alignment chevron needs to operate on the full diameter, i.e., if the portion of the cylinder that contacts the chevron had half the diameter, the chevron would be correspondingly smaller. Relatively small groove and pad kinematic mounts can be accurate to better than a few microradians.

One issue that you might consider is that if there is a lot of relative movement and dragging, the long term surface wear might be an issue. A plastic against steel will lose material over time.

TTFN
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RE: How shallow an angle of V would still align with a round metal bar?

The simplicity and familiarity of the above solution got me thinking I might have overcomplicated things.  I had.  Enormously.  If we tip the problem upside down it can be reduced almost to a classic text book case that any high school physics student could solve.  See attached diagram.

The critical value for alpha, the value at which the bar is just at the point of sliding along the vee (or, un-inverting, the vee is just at the point of sliding down the bar) can readily be seen to be given by
alphacrit = arctan(1/u)
where u is the coefficient of friction.  For values of alpha greater than this critical value sliding will not occur.

RE: How shallow an angle of V would still align with a round metal bar?

Denial, see previous posts relating friction coefficient to angle.

Ted

RE: How shallow an angle of V would still align with a round metal bar?

Yair, Hydtools, you got there directly.  And week earlier.  I took a (very) long way around, unable to see past what I took to be a complicated geometry.  But I had a bit of fun along the way.

RE: How shallow an angle of V would still align with a round metal bar?

It was an interesting exercise to analyze the rotation.

Ted

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