# Center of Gravity and a Balanced TV Stand

## Main Question or Discussion Point

I am building a stand for my LCD/LED TV and Mount. See several pics below: Calculation Sheet, Mount Diagrams, and Side View of the TV. The Mount will be hanging on 2 poles, and the TV hangs on the front of the Mount.

Am I computing the Center of Gravity correctly? in order to determine where to let the whole connected device hang, so it will be centered on my stand, distributing the weight at the center?

I show my Calculation Sheet in one of the attachments, but basically the final formula filled in with weights and distances is:

Center of Gravity = (4.75 in)x(36 lbs) + (10 15/16 in)x(49.8 lbs) / (36 + 49.8) lbs = 8.341 in

where

4.75 and 36 are the Center of the Mount and Mount Weight,
and
10 15/16 and 49.8 are the Center of the TV and TV Weight,

Length measurements are taken by setting the back of the Mount as 0. Then, later translating the interval to the back of the Stand.

The more I think about it, does the point of weight also have to do with the lever action of the 2 poles holding the weight up, on the stand? I also decided to let the poles rest on the ground, after passing them through my stand ... so this is really more about balancing the whole, so there is the least chance of it toppling over.

Please discuss and help me determine a best position for the 2, 2" holes for the poles.

Currently, I calculate 6.159 in. (center of poles being 3 in. from the back of the Mount) from the back edge of the 23 in. deep Stand, will place the Center of Gravity at 8.341 inches of the Mount and TV, over the center of the stand (from back to front).

Does this make sense?

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nvn
Homework Helper
billycar: Does your mount weight (36 lbf) include the weight of your two vertical poles? If not, include the weight of the two vertical poles in your center of gravity (CG) calculation.

What is the weight of the stand?

No, currently it does not. As soon as I find 2" OD metal poles, I'll weigh them, and include them in the scheme. Any help locating some would also be appreciated ... looks like the closest (but No Cigar) metal pipe at the Plumbers is ID 1.5 and OD 1.9. That includes PVC pipe as well. Evidently, "tubing" (vrs "pipe") is measured to OD. So far, haven't located anything short of ordering it from the Mount Manufacturer's Dealer, and they are \$125 (plus Shipping) for 2 - 6 ft lengths ... : ( ... may need to surrender and do so.

I did read recently that Stripper Poles are usually 2 inches in diameter ... : ) ... but I'm sure they go for a premium, and are probably too slick ...

Looking for a way to weigh the stand ... it is a modified TV Stand from a local DAVs, and measures (w x d x h) 33 x 23 x 24 (including the casters), and with the additional hardware and backing to strengthen it, has become a challenge to manipulate ... which is what I wanted.

Should I be computing the Center of Gravity of all the components (Poles, Mount, TV and Stand)? and move the Pole/Mount/TV part, until the point is in the middle of the stand? Width-wise, due to symmetry, I assume it is along the centerline of the Stand; so, I am concerned about front to back.

I was just considering a horizontal line through the Mount and TV ... the poles come through the Mount 3 in. from the back of the Mount ...

Other Considerations

The Mount-ed TV does tilt 10º down and 35º up, so maybe the "thickness" of the TV needs to be modified ... or to simplify things, just do the calculation for an at rest vertical alignment. It does pivot along its centerline.

Since the casters are not at the true edge of the Stand, I need to remember to measure the focal point from this offset as well ...

Onward ...

nvn
Homework Helper
All we need is an estimate of the weight of the vertical poles. It sounds like you will use steel tubes, right? Or will they be aluminum? The OD is 50.8 mm (2.0 in). What will the tube wall thickness be? Just estimate or guess. And what is the length of each vertical tube? If you give us these dimensions, we can compute the weight of the vertical poles, because steel has a density of 7850 kg/m^3.

Just post a rough estimate or guess for the stand weight. The estimate of the stand weight does not need to be accurate. Is the stand weight approximately uniformly distributed? Or is the back much heavier than the front? You can actually weigh the front and back separately, by tilting the unit up on a scale, and weighing the front and back separately.

The Premier Mount poles I'm considering are 72 in. tall and 2 in. OD. They say the Shipping weight for Chrome ones are 21 lbs (for Black ones, 22 lbs):

http://www.premiermounts.com/product.asp?partid=93 [Broken]

That's for 2 of them.

My reinforced Stand now weighs ~ 90 lbs.

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nvn
Homework Helper
billycar: For now, recompute your CG, as mentioned in post 2, including the weight of the vertical poles, but excluding the stand weight. Use the same method you used previously. The answer will slightly differ from your previous answer, 8.341.

If the vertical poles are steel, and if we are to believe the shipping weight, then it means the vertical tubes have a wall thickness of 2.11 mm; and therefore, the total weight of both vertical poles is 90.90 N (20.44 lbf).

nvn
Homework Helper
billycar: Notice, your CG location now becomes 185.77 mm (7.314 in).

If the vertical poles did not rest on the ground, you would want the CG exactly over the center of the stand, in which case the centerline of the vertical poles needs to be 182.53 mm (7.186 in) from the back edge of the stand.

However, now that you decided to let the vertical poles fully rest on the ground, it no longer matters where the vertical poles are located. Interestingly, no matter where the vertical poles are located, you always get the same reaction forces on the caster wheels. Therefore, the best location for the vertical poles would be 182.53 mm from the back edge of the stand.

I get ... (using Shipping Weights of the Poles):

CG = {(4.75 in)x(36 lbs) + (10 15/16 in)x(49.8 lbs) + (3.0 in)x(21 lbs)} / (36 + 49.8 + 21) lbs = 7.291 in

where 3.0 in and 21 lbs are for the Center and (Shipping) Weight of Chrome Poles

For the Black Poles, the CG = 7.251 in.

*****

Interesting what you say about the poles at rest on the floor. A friend suggested I consider attaching the poles onto the back of the Stand, since it is now reinforced, letting them rest on the floor.

Same reaction forces since the poles are perpendicular to the Stand? Would that hold if the poles were behind the wheels, but resting on the floor?

I don't think I would need to worry about back tilt, since the Stand's weight is greater than the hanging objects? Actually, in that location, I believe I would have the least chance of backward tilt (due to the weight of the Stand), and least forward tilt chance, since the distance from poles to front wheels is greatest.

Maybe attaching the poles to the back, and resting on the floor, is a best solution?

Or would you still want to have the CG of the Poles/Mount/TV or Mount/TV, located between the casters?

So, for my calculations for the Black Poles, have the CG of 7.251 in., measured in from a 0 point directly above the back caster.

nvn
Homework Helper
In that location, I believe I would have the least chance of backward tilt ...
That is incorrect. Also, 7.251 is measured from the stand back edge, not from the wheels.

Although in post 7 I said you always get the same wheel reaction forces no matter where the vertical poles are located, that statement is somewhat irrelevant, because that statement is true only when no horizontal tipping force is applied. Therefore, see the following paragraph, instead.

If the vertical poles are located at 7.251 inch from the stand back edge, and you apply a rearward horizontal force to the assembly to cause the front wheels to just start to lift off the ground, this horizontal tipping (overturning) force is ~1.43 times greater than the tipping force when the vertical poles are located at the rear wheels. Therefore, the greatest stability occurs when the vertical poles are located at 7.251 inch from the back edge of the stand.

What if the 2 poles were on the back of the Stand, behind the wheels, resting on the floor? That would make the center of the poles ~5.5 in. behind the wheels (about 1+ in. from the back, since the wheels are located ~4.5 in. from the back surface towards the center).

What type of stability concerns would I have then? as compared to placing the center of the Poles at a distance CG from the back towards the center of the Stand?

Do we need to consider the center of the Poles, as some kind of Center of Gravity focal point of the load in some sense? since the Poles/Mount/TV form a solid object? I'm trying to understand the mechanics of it, being somewhat of a layman. When you suggest I place that point at 7.251 in. from the back, I wonder why choose that point on the Poles.

I appreciate your attention to the issues here, and your patience with my questions.

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nvn
Homework Helper
Oops. In post 10, I got mixed up on one of your numbers and quoted the wrong value. In post 10, change each occurrence of 7.251 to 7.209. From now on, let's always use 21 lbf for the weight of the poles, regardless of type. Therefore, the CG becomes 7.291 inch.

I am surprised your wheel inset is so large. If you are sure that is correct, I will go ahead and change my numbers. This radically changes the results, as follows.

If the poles can slip axially, then placing the poles 1 inch behind the stand back edge gives you about the same (low) stability as placing the poles anywhere else, such as at the rear wheels. However, it is doubtful that the poles can completely slip freely during tipping. Therefore, below, I assume the poles cannot slip axially.

If the poles, resting on the ground, are located at a distance xp forward of the stand back edge, and you apply a rearward horizontal force to the assembly to cause the front wheels to just start to lift off the ground, the relative horizontal tipping force, F1(xp), is listed below, where F1(xp) means F1 as a function of xp. Positive xp means the pole centerline is forward of the stand back edge; negative xp means the pole centerline is behind the stand back edge.

F1(8.418) = 1.147
F1(7.209) = 1.266
F1(6.000) = 1.147
F1(4.500) = 1.000
F1(3.000) = 1.124
F1(0.000) = 1.372
F1(-1.00) = 1.455
F1(-2.00) = 1.538​

The above values (to the right-hand side of the equals sign) are dimensionless because they have been divided by the minimum tipping force, which occurs when xp = 4.500 inch. In this way, you can think of the above listed values as the stability. Notice, the minimum stability occurs when the poles are located at the rear wheels. The maximum stability, say 1.538, occurs when the poles are 2.00 inch behind the stand back edge. The stability reaches another local maximum when the poles are 7.209 inch forward of the stand back edge. If you move the poles further forward than 7.209 inch, the stability decreases, because the assembly could then tip forward in an identical way as listed above. The above function is linear; therefore, you can use linear interpolation between any of the above values (or linear extrapolation).

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Regarding the position of the 2 1/2" casters: they are of the ball bearing type, mounted on 3" x 4" metal plates, which are bolted to the base of the Stand ... actually, on the front ones, the center of the plate is 3.5" from the front, and after adding a 1/2" backing, the back ones are 4" from the back wall. Apologies for the changing numbers ... at times, I work late into the night, and may overlook something here and there ...

I'm still thinking I should compute the CG of the Poles/Mount/TV, when I get a final weigh-in of the components, and then place the center of the Poles at the location which makes that CG fall perpendicular to the center line of the casters, from front to back; but, I am also trying to accept that the force at that CG is acting at the center point of the Poles, when they are attached to the Stand, so, the Poles should be positioned at the CG point, but still measured between the caster locations.

Assuming the weight distribution of the hardware is symmetrical from left to right, I should place the 2 poles, symmetrically about the center of the stand, from left to right.

Once setup, the poles will be stationary, except they may be raised slightly, when moving the setup about the Studio. Otherwise, they rest against the floor, in a locked down position.

A couple of other considerations will be the Tilt Movement of the TV: 10º upward, and 35º downward about its centerline (probably only making the TV virtually thicker for our calculations ... although its centerline should remain constant through this motion).

I will also be able to rotate the TV 90º to Portrait and Landscape orientations (I don't think this affects the calculations). To simplify the problem description, I decided to hold off on this until now.

nvn
Homework Helper
billycar: The screen tilt and orientation do not matter, because the CG does not change appreciably. Secondly, the force at the center of the poles is already taken into account in the F1(xp) values. I updated the F1(xp) values, below, based on post 13.

F1(8.418) = 1.188
F1(7.209) = 1.302
F1(6.000) = 1.188
F1(4.000) = 1.000
F1(2.000) = 1.159
F1(0.000) = 1.318
F1(-1.00) = 1.397
F1(-2.00) = 1.476​

Place the centerline of the poles at the xp location having the highest stability. The xp location is the number in parentheses, above. The stability is the number to the right-hand side of the equals sign. Notice, there are two maxima. One at xp = 7.209 inch, and one at say xp = -2.0 inch. You can pick the stability you prefer. The xp value is measured from the stand back edge to the pole centerline.

What is the function you are using to compute stability? Linear F1(xp) = m * xp + b? Where does it come from?

My poles should arrive next week some time, and I will be able to finish construction of my Stand, taking actual weights, and computing CG again. Probably won't be too much different than what we have come up with so far.

Thanks again for all the comments.

nvn
Homework Helper
That is correct, m*xp + b. The functions are listed below, but it would be too time-consuming for me to explain their derivation on here.

F1(xp) = 0.09424*xp + 0.6230, for 4.0 ≤ xp ≤ 7.209.
F1(xp) = -0.07942*xp + 1.318, for xp < 4.0.

Well, I made another error ... the weight of the Stand is actually 45 lbs. I've added 2 more support layers to the top and bottom, and decided to weigh again. My apologies. I must have been trying to eyeball the scale, from an awkward angle, and read 90 (kg) and thought it was (lbs).

After reading through the discussion, I decided to also weigh the front and back of the Stand. Front and Back weights are 13 lbs and 21 lbs (wheels of opposite side resting on the table).

Then, lately, not having much feel for weight of objects, factors in to the error making.

Do you have a website location that might explain the derivation of the "tilt force" function?

nvn
Homework Helper
I don't know of a good web site. You would basically want to take an engineering course called Vector Mechanics: https://oli.web.cmu.edu/openlearning/forstudents/ [Broken]. Tilting the stand while weighing the front and back really changed the total weight. If you get a chance, weigh the front and back (at the wheels) when the stand is level (by placing the other two wheels on a block). Turn all wheels out sideways, so they will be exactly at the 4 inch and 3.5 inch wheel locations during weighing. When the stand is level, I think the front plus back weight should equal the stand total weight. Once you do this, we could pinpoint the stand CG. And then I will update my numbers.

Also, I have currently been assuming the overturning force (F1) is applied horizontally at the screen center. And I have been placing F1 at a height of 60 inch above the bottom surface of the stand. Decide if this sounds like the worst scenario. The greater the height of F1, the more easily the assembly is overturned.

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The metal Poles finally arrived, and now I have accurate weights for them ... and I have also re-checked my weights for the other components (and now am using the correct scale of lbs). My scale has both lbs and kgms on it, and earlier, reading it upside down, I picked the larger numbers, which are the kgms ... big apology ... my eyesight isn't the best. (1)

In response to the last message above: even though the poles are 72 in. high, I would not go much over 60 inches. I would want the center of the screen, landscape and portrait, no higher than eye level (5' 6"), with me standing. The bottom of the Stand is ~3.5" off the ground. Perhaps say 66" maximum height for center of the Mount and TV.

Actual weights are:

Mount = 24 lbs (2)
Stand = 105 lbs (3)
Stand Front = 45 lbs (4)
Stand Back = 60 lbs
Poles = 18 lbs (5)
HDTV = 49.8 (6)

I needed to add 4 lbs of PVC pipe and Pipe Pressure Clamping Rings, that the metal tubing will slide down into on the Stand. This weight is now included in the Stand at the 7" mark from the back.

Now CG Mount/TV/Poles = 7.7635 = (4.75 x 24. + 10.9375 x 49.8 + 3.0 x 18.0) / (24. + 49.8 + 18.0)

As an aside ... if I compute the CG of the Stand = 10.6786 = (5.75 x 60 + 17.25 x 45) / (60 + 45), what is wrong with lining up (plumb line) the CG of the Mount/TV/Poles with the CG Stand, and calling the whole balanced?

I would say that the poles would need to be located at 5.9151 in. = 10.6786 - (7.7635 - 3) from the rear, which would make this appear (to me) to occur.

Notes

(1) I made an error in my weight measurements (my apologies), and decided to wait until I got a more precise measurement on the pole weight, before asking the Helper to proceed ... For the Poles, I decided the "Ship Weight" was probably way off, if it was similar to the Mount's Ship Weight of 36 lbs, Actual Weight 24 lbs.
(2) (yes, evidently, the packaging and extra hardware / bolts weighs 10+ lbs). I will use 4 bolts and some small plates to attach the HDTV to the Mount, and did not include that weight here.
(3) Completed Stand with finished design elements included: Poles Mount, Shelf
(4) wheels level front and back, and turned outward from center
(5) Poles (Shipping Weight) = 21 lbs
(6) Using the published HDTV Weight

nvn
Homework Helper
billycar: Your mount/TV/poles CG is correct. I do not yet understand the wheel locations. Last I recall, the rear wheels were 4.0 inch from the stand back edge; and the front wheels were 3.5 inch from the stand front edge. Therefore, I did not yet understand your stand CG calculation, because it appears to use a wheel inset of 5.75 inch for all four wheels.

The larger numeric values on a scale would be lbf; the smaller numeric values would be kg. The quantity of lbf would be 2.2 times the quantity of kg.

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I divided the Stand into 2 equal sections, assigning 60 lbs to the back, and 45 lbs to the front. 1/2 way into the back block is 5.75 in and 1/2 way further into the front block is 17.25 inches from the back.

____Back_________Front
__________________________
|___________ |_____________|
|___________ |_____________|
|____60_____ |______45_____| Back and Front Weights, Both Wheels at 3 inches from Edge
|___________ |_____________|
|____4.0_____|______3.5____| Distance to Center of Swivel on Plate where attached to Stand
_____/ /_____________\ \
____O______________O _____ Back Wheel is Swivel w/ Locking Mechanism; Front Wheel is Swivel - Non-locking
|...3..|________________|..3...| Distance to Center of Wheels on Scale (from Outer Edge of Stand)

Regarding the lbs vrs kgms, by larger, I meant the Font Size of the numerals were larger for the kgms scale numbers than for the lbs scale numbers.

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Thank You. Looks like we responded at the same time. Any difference in your calculations, looking at the clarification of wheel location?

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Thank You so very much for all your time and considerations.

nvn
Homework Helper
The stand CG is actually, (3.0*60 + 20.0*45)/(60 + 45) = 10.286.

I am changing the nomenclature, slightly. Below, I show that I am dividing the horizontal tipping force, F1(xp), by the minimum tipping force, F1min, to obtain the relative stability. The value of F1min is F1min = 17.42 lbf. Multiply the right-hand side of the following list by F1min to obtain the actual tipping force, F1(xp). F1(xp) is applied horizontally at a height of 63 inch above the stand bottom surface, in the front-to-back direction.

F1(11.00)/F1min = -1.194
F1(10.00)/F1min = -1.278
F1(9.000)/F1min = -1.362
F1(8.660)/F1min = 1.390
F1(8.000)/F1min = 1.335
F1(7.000)/F1min = 1.251
F1(6.000)/F1min = 1.167
F1(5.522)/F1min = 1.127
F1(5.000)/F1min = 1.084
F1(4.000)/F1min = 1.000
F1(2.000)/F1min = 1.191
F1(0.000)/F1min = 1.383
F1(-1.19)/F1min = 1.497​

The xp location is the number in parentheses, above. The stability is the number to the right-hand side of the equals sign. Notice, there are two maxima. One at xp = 8.660 inch, and one at xp = -1.19 inch. You can pick the stability you prefer. The xp value is measured from the stand back edge to the pole centerline. When xp = 5.522 inch, the mount/TV/poles CG is over the stand CG. You can use linear interpolation between any of the above values (or linear extrapolation).

Footnote: Even though the wheels were turned forward and aft for weighing of the stand, the wheels are all turned sideways in the above stability calculations, which is a worse condition for stability. I.e., the rear wheels are inset 4.0 inch, and the front wheels are inset 3.5 inch.

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Points of Clarification Needed Deciding Where to Locate the Pipe Mounts

To place the CG of the Mount/TV/Poles over the CG Stand:

Location of Center of Poles = CG Stand - (CG Mount/TV/Poles - Distance from Back of Mount to Center of Poles)
________________________= 10.2857 - (7.7635 - 3.0000)
________________________= 5.5222

And this is measured from the back edge of the Stand, even though we measured from the 3.0 inch shift inward due to the wheels, since the CG Stand comes from the distribution of the weight of the Stand from Back to Front. (It is coincidental that the distance from the back of the Mount to the center of the Poles is also 3.0 inches.)

Suppose I mounted the poles to the back of the Stand:

Using my 2" I.D. PVC Pipe Mounts (2 3/8 O.D.), the Center of the 2" O.D. metal Poles would then be at -1 3/16 = -1.1875.

I would have greater relative stability from approximately 1.2 (at 5.5222) to 1.5 (at -1.1875). By stability, we are talking about tipping from front to back, and back to front?

It would make more sense to mount the PVC Pipes w/Poles on the back of the Stand?

Or is it better, in some sense, to have the PVC Pipes inside the Stand, contained within the spread of the wheels? perhaps at the new value above of 8.660?

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