Assuming it is a steady state acceleration (or ramp from 0 very slowly up to the final constant acceleration) with no transient sloshing around, then you can solve it as follows.

1 - For the stationary case (gravity force but no horizontal acceleration), determine what angle could you tilt the glass without spilling.
2 - What level of horizontal acceleration when vectorially combined with gravity "acceleration" will give a total acceleration in a direction that satisfies the angle of tilt calculated in part 1.

=====================================
(2B)+(2B)' ?

acceleration applied at CG of fluid.

this isn't for work, right, so discouraged at the "for work" site.

another day in paradise, or is paradise one day closer ?

No, I need to find out max acceleration on my conveyor mechanism.

Years of spilling stuff makes me think that "jerk" may be more important than acceleration.

Jerk is a form of acceleration, isn't it? I am trying to figure out max acceleration to use in my specs.

jerk is the time derivative of acceleration (d3x/dt3)

why do your specs care about water spilling out of a polystyrene cup ? unless you've calibrated your actual operation to water in a coffee cup ...

another day in paradise, or is paradise one day closer ?

Have you considered height of fluid compared to height of cup? Have you considered frequency range (F-min. and F-max.)? A quick experiment on my half-filled (or half-empty) coffee mug showed that tapping/impacting with my finger knuckle (closed fist) produced minor sloshing as compared to slight shaking of cup with my hand that produced considerable sloshing. I believe the g-level was much higher in the first case. I could measure it! Fluid in cup or tank has a sloshing natural frequency, so the frequency range of the applied acceleration along with amplitude should be considered.
https://www.bing.com/search?q=tak+fluid+sloshing+n...

Get a cup of coffee,
Walt

Often, spillage is from sloshing, which is non-trivial to calculate

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What if I suddenly wanted to stop my conveyor? Wouldn't the speed and acceleration be detrimental to prevent spillage?

The same rationale applies. The change in direction of the jerk simply sloshes water in the opposite direction

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Are you actually transporting half full cups of water or just using that as an example.

Maybe look at what sushi restaurants do for their revolving conveyor of food....

Or restaurants when you place the tray on a moving conveyor.

I think though this is the sort of thing that needs some experimentation as there seem to be many variables involved (weight of the cup, friction, how full the cup is, whether a truly "smooth start / stop can be applied ( see flywheel issue below) final speed of the conveyor etc)

You probably need a big flywheel on the motor drive so that it can't accelerate or decelerate very quickly.

How fast does the conveyor go when it reaches max speed?

Is there an issue in doing this ramp up / down over 10 seconds, or 20 or 30?

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If you know the static level of water and dimensions of the glass you can calculate the acceleration at which the water level reaches the brim.

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for a steady acceleration ...
and the taper of the cup is going to mess things some (but nothing a bit of math won't solve)

another day in paradise, or is paradise one day closer ?

yeah I am working on it now, but the calculations are taking much much longer as the shape of the glass is not as simple as a rectangular tank or cylinder.

I'm getting 2.75L for the cup and only 0.5L for water, so less than half full?

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As has been said, for this Jerk is what matters (rate change of acceleration)
Yes you will have a max acceleration, but how quickly it changes matters more in this case.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

I got a_x=10.26 m/s^2. But somebody should check.

for the cup volume I get .7L ... average 3" dia, 6" high = 42in3 = 700cc

what constrains the water to have the same CL when deflected ? is the volume above = the volume below the original water level ? doesn't look like it (if the center stays in the same place.

we know the initial freeboard, we know the initial empty volume of the cup. This is still the empty volume with the water tilted.

another day in paradise, or is paradise one day closer ?

goutam_freelance , great work! I was able to calculate the height of the water, but nothing more. Could you please explain how you drove the equation you used to solve for tan (alpha)?

Initial volume=Volume below lowest point on water surface+1/2(Full glass volume-volume below lowest water surface).
Water surface divides the volume above lowest surface point in 2 equal halves.

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As I told before the concept needs to be checked. I shall check again.

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I think the glass will fall over before the nominal water level sloshes. I don't see a weight/CG for the material that forms the cup, so there's no way to tell where the water is to begin with.

The simple method first-order acceleration test is to put a glass on a slope of the same material as the conveyor belt and tilt it until it falls over.

I think you're right 3DD. there is 124mm of water, so the CG is somewhere north of 62mm, and the base is radius of 25mm. so a combined acceleration vector of nearly 45deg (10m/s2 down and 10 m/s2 to the right) would be considerably outside the base ...

another day in paradise, or is paradise one day closer ?

On further thought I found that the procedure given by me is approximately true, since the glass side angle is small( 9 deg approx). The formula is correct if the glass is perfectly cylindrical. But till now I am unable to find a formula for division of cone volume by a diagonal plane. Hope somebody here will be able to help.

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The fun thing about a diagonal cut on a cone is it forms an ellipse. While there is a formula to calculate the area of an ellipse, AFAIK there is no formula to exactly calculate the length of the ellipse perimeter except for a circle.

https://www.mathsisfun.com/geometry/ellipse-perime...

With a cylinder this is a trivial calculation; on a cone, not so much.

3DDave, I have already figured that part of the calculation. I know the max acceleration that would cause the glass to tip over on its edge. I was trying to figure if the water spillage would happen at smaller acceleration.

goutam_freelance, I am still having a hard time to understand how that equation gives you volume. How is water density not part of that calculation? I am very confused. Could you please elaborate that equation?

Yet again my not asking for a photograph leads to misunderstanding.

Water density doesn't matter if all you are concerned about is where the liquid reaches the lip of the cup. All liquids will have a surface normal to the local acceleration.

So, since those dimensions must be the inside of the cup you should be able tip the cup so the water just reaches the lip and the tangent of that angle is the ratio of horizontal acceleration to vertical acceleration that will reach that point.

Yeah, I realized that after engaging with you guys and reading up some more on that. But I am still curious how to calculate it from theoretical point of view. Goutam_freelance seems like is onto something, but I am not really sure how he came up with that equation where he finds tan(alpa). Years of not doing much hand calculations and relying on softwares must have dumbed me down. Are you able to see what he has done there?

rather than looking at the ideal case of a steady acceleration and setting the water surface to the lip of the cup (and yes we haven't solved the volume of a cone cut by an inclined plane ... maybe do it in CAD ?) ...
I think the critical case was identified by 3DDave ... when does the acceleration vector become unstable ... move beyond the footprint of the cup. If we have an initial water level of 124mm then the CG height is roughly 62mm, and the base radius is 25mm. so an acceleration of 25/62, approx .5g, would be close to tip over the cup. Yes, I know it'll be more than that as we need the contact pressure at the other side of the cup to go to zero (from the static pressure). But this seems to be much lower than the estimates of the water surface (which seem to be about 1g).

another day in paradise, or is paradise one day closer ?

This is pretty simple to set up in CAD.
Simply vary the angle until the volume becomes 500cc (same as 0.5kg of water)
Angle is apx 28.5 degrees



Now find the CG (another CAD feature) and drop a line perpendicular to the fluid surface.



CG line is significantly to the left of the cup edge, therefore the cup will tip over before it spills. However, if the cup were somehow glued to the conveyor, the acceleration is tan(28.5)g

@hungry_kid
Lately I have some free time so I decided to' Go back to School'.

The following are from Schaum's Outlines-Fluid Mechanics:


Now refer the original figure posted by me.


@handleman, thank you for the info. Yes it can be done with CAD. But I was looking for a formula to put it into an equation.

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Formula for the volume of a cone truncated at an angle is here:

https://keisan.casio.com/exec/system/1483681473176...

This problem could be solved geometrically/algebraically. The water surface should remain approximately normal to the resultant acceleration vector (gravity + local acceleration). Setting up a system of equations should make this problem trivial for analytical solution. I'd set it up as follows: volume of empty glass - volume of air in glass (inverse of liquid). The volume of air is the integral of a set of ellipses through an angle which describes the surface of the liquid. The angle (which is the upper bound of the integral) would be normal to the resultant acceleration vector. The rest is algebra. In this way one could develop a set of equations useful for resolving with different sized containers and different amount of liquid.

@handleman, thank you. The formula you gave was very useful. I have redone the calculation. The angle and hence acceleration tallies with CAD output.
The numerical solution was done in Matlab using fzero() function.

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You guys are awesome! Thanks a lot

I'd take your 'glass' to anywhere that has a conveyor driven by a VFD (lots of places use them) and just test it. You may see things and results that you haven't even considered and you will know for sure.

Keith Cress
kcress - http://www.flaminsystems.com

I am not convinced that a static geometric model is representative of the water sloshing, especially without knowing the force-vs-time for the conveyor start and stop events. It might make a difference if the spillage is defined as one drop or some percentage of the volume in the cup. Some testing could provide the solution or at least validate the model. I am with itsmoked on this one.

Walt

1.  Use ElectricPete's 14Jan21@14:53 approach to find the surface's "steady state" slope under a constant acceleration.
2.  Double that slope to get a reasonably accurate estimate of the maximum transient slope that would be caused if the acceleration came on instantaneously.

[Correction/edit at 20:30 18Jan21 UTC.]  It is, of course, the tan() of the slope that should be doubled.

Starting instantaneously and "jerking" are simply very high bursts of acceleration.

This analysis seems to assume the cup stays flat on the surface at all times which seems a little unlikely.

We also have no steady state speed.

As well as the acceleration there will be a deceleration and also a reduction in acceleration to zero to contend with when you get close to your final speed.

All these are difficult to judge without some experiments

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Also: If you get a response it's polite to respond to it.