Calibrate Scale: Sliding Weights Not Springs

[Y.M.E.]

I have a scale, the kind for weighing people, and I'm trying to calibrate it. It uses sliding weights, not springs. You step on the scale and then slide the two weights along the beam until it balances. Suppose I keep the smaller weight at zero and slide the larger weight to the 50-pound mark, but it requires 51 pounds on the scale to make the beam balance. This means that the sliding weight is too heavy, so some material needs to be removed from it. I figure that its new weight ought to be 50/51 times its original weight. Conversely, if 49 pounds on the scale causes the beam to balance, then the sliding weight is too light, and enough material needs to be added to it to make its new weight equal to 50/49 times its original weight. Is this correct?

The reason I ask is that I adjusted its weight in this way, but I ended up overshooting. Initially, the scale read too high, but after being adjusted, it was not correct, but, rather, it read too low, although it was off by a smaller amount than before. So, I changed the weight again, in the other direction this time, using the same rule, and overshot again. And likewise a third time. So clearly I'm doing something wrong here, but I can't figure out what.

 

[Doug Huffman]

There is often a screw mounted counterweight opposite the fulcrum from the adjustable balance weights.

 

[Y.M.E.]

There is often a screw mounted counterweight opposite the fulcrum from the adjustable balance weights.

Yes, my scale has that. It's for zeroing the scale. That is, you're supposed to set it so that the beam is balanced when the sliding weights are both at zero and no weight is on the scale, which is what I did. But if, after that's done, the scale reads the wrong weight when something is placed on it, the way to fix this, different, problem is to change how heavy the sliding weights are. My question is, by how much should their weight be changed?

 

[Doug Huffman]

Sorry that I will not be of further help. I have never seen a selection of steelyard or triple beam balance weights that would suggest that as a proper calibration. Might the fulcrum(s) have shifted?

 

[jack action]

How sure are you that the weight you put on the scale really weighs 51 lb?

How sure are you that the weight you added on the beam is 1/50th of the original weight?

 

[Y.M.E.]

How sure are you that the weight you put on the scale really weighs 51 lb?

How sure are you that the weight you added on the beam is 1/50th of the original weight?

So, the rule I used for adjusting the sliding weight is correct? And that's why you're looking for the source of the problem elsewhere?

 

[jack action]

So, the rule I used for adjusting the sliding weight is correct? And that's why you're looking for the source of the problem elsewhere?

Yes.

 

[Danger]

I honestly can't think of anything to account for it other than what Doug mentioned; your fulcrum must be misplaced. One way to tell: is it a percentage inaccuracy or a constant weight discrepancy?

 

[Y.M.E.]

I honestly can't think of anything to account for it other than what Doug mentioned; your fulcrum must be misplaced. One way to tell: is it a percentage inaccuracy or a constant weight discrepancy?

I don't understand. Wouldn't a misplaced fulcrum or a sliding balance weight that's too heavy or too light both cause a percentage inaccuracy? Not zeroing the scale via the movable counterweight would cause a constant weight discrepancy, I think, but I did zero it.

The scale seems to have more friction than it should. The beam doesn't oscillate much after stepping on the scale, and if I gently bounce up and down while standing on it, the beam often settles down to a slightly different position. I've taken it apart, and some places that I think need knife edges aren't very sharp at all. (They don't look like they used to be sharp but just wore down. They look like they never were sharp to begin with, which really surprised me.) I didn't think that the effect of all this would be large enough to account for what I described in my first post, which is why I asked the question, but maybe it was.

 

[Y.M.E.]

How sure are you that the weight you put on the scale really weighs 51 lb?

How sure are you that the weight you added on the beam is 1/50th of the original weight?

Pretty sure.

I have a Roberval balance that's graduated in tenths of a gram, although parallax makes it hard to read the pointer that precisely. So, let's say +- 0.2 g. It can weigh up to 210 g using the pair of sliding weights that are part of it, and more than that if I place known weights on the other pan. I have two 200-g weights and one 100-g weight with 30 mg and 16 mg tolerances, respectively (OIML class M2), and the balance gives the correct reading for them.

I used the balance to weigh the larger sliding weight of the big scale, and it weighs about 280 g.

Most of the 50-lb weight that I put on the big scale was nine reams of paper, 500 sheets each. On the small balance, I weighed 110 sheets and, separately, the paper wrapping that encloses each ream, then extrapolated to nine reams. The rest of the 50 lbs was also weighed on the small balance.

 

[Doug Huffman]

LOL Learn the differences among accuracy, precision, indication and repeatability.

I occasionally help a local fisherman-retired. We repack fish to 50# with 20# of ice using a balance scale so old that I have to peer at the indicator to see its rulings.

It works fine, has lasted a loong time, don't rust (any more), bust or collect dust. It fails safe and drains to the bilge.

 

[Y.M.E.]

LOL Learn the differences among accuracy, precision, indication and repeatability.

How did I use them incorrectly?

 

[Doug Huffman]

A pointer is properly only read to half of the smallest division. Imagine teaching a non-technical person to read a log-scale indicator.

Sorry, I am not a metrologist.

 

[Y.M.E.]

A pointer is properly only read to half of the smallest division.

Ok. But I don't see where I did otherwise. The smallest division on my balance is 0.1 g, and I increased that to 0.2 g to allow for the difficulty of reading it.

 

[jack action]

Another thing, when you add your weight on the sliding weight, you need NOT to change the CG position. If you don't keep the CG location at the same place, you are effectively changing the lever arm length at the same time.

Even if it is at the same place with respect to the beam length direction, if it is not at the same place in the direction perpendicular to it, when the beam start to move down the lever arm start to changes and the more it goes down (or up), the more it changes, making it impossible to return.

I don't know how much precision is required on CG location for your particular balance, but that could be something to investigate.

 

[Y.M.E.]

Another thing, when you add your weight on the sliding weight, you need NOT to change the CG position. If you don't keep the CG location at the same place, you are effectively changing the lever arm length at the same time.

This did occur to me, and I also worried about the fact that the sliding weight hangs from the beam in two places, not only one. But then after thinking about it some more, I concluded that re-zeroing the scale after adding to the sliding weight would take care of these things. Do you agree?

Even if it is at the same place with respect to the beam length direction, if it is not at the same place in the direction perpendicular to it, when the beam start to move down the lever arm start to changes and the more it goes down (or up), the more it changes, making it impossible to return.

I didn't think of that before, but now that you mention it, I agree. Raising the center of gravity of the beam increases the sensitivity of the scale, and if it's raised above the pivot, the beam will tend to rotate all the way one way or the other.

 

[jack action]

This did occur to me, and I also worried about the fact that the sliding weight hangs from the beam in two places, not only one. But then after thinking about it some more, I concluded that re-zeroing the scale after adding to the sliding weight would take care of these things. Do you agree?

I don't think so. Here's why:

You got $m_1$ that is your 51 lb, $m_2$ that is your 280g and their respective lengths from the fulcrum, $L_1$ and $L_2$, such that:
$$m_1 = m_2 \frac{L_2}{L_1}$$
Then you accidentally increase your length by $L_a$ when adding the additional mass $m_a$:
$$m_1 = \left( m_2 + m_a \right) \frac{L_2 + L_a}{L_1}$$
So you decide to zeroing your scale again. Now $m_1$ is replace by $m_0$ (the weight of the scale tray without the 51 lb) and $L_2$ by $L_0$:
$$m_0 = \left( m_2 + m_a \right) \frac{L_0 + L_a}{L_1}$$
You then zero it by increasing the lever arm by $L_x$ (could be negative, if decreased):
$$m_0 = \left( m_2 + m_a \right) \frac{L_0 + L_a + L_x}{L_1}$$
Then you go back to your 51 lb measurement:
$$m_1 = \left( m_2 + m_a \right) \frac{L_2 + L_a + L_x}{L_1}$$
For this to respect your first calculation the following has to be true:
$$\frac{L_2}{L_1} = \frac{L_2 + L_a + L_x}{L_1}$$
Which means that:
$$L_a = - L_x$$
Going back to the zeroing equation:
$$m_0 = \left( m_2 + m_a \right) \frac{L_0 + \left(- L_x\right) + L_x}{L_1} \\ m_0 = \left( m_2 + m_a \right) \frac{L_0}{L_1}$$
Which is impossible since we already know that:
$$m_0 = m_2 \frac{L_0}{L_1}$$

But I did this quickly, maybe I overlooked something.

 

[Y.M.E.]

I don't think so. Here's why:
[...]
But I did this quickly, maybe I overlooked something.

I'm having a bit of trouble following that.

Maybe it would be easier if we try to separate the effects of adding weight and shifting the center of gravity. We want to add weight; we don't want to shift the center of gravity (unless it doesn't matter anyway).

So, suppose we keep the weight the same and just shift the center of gravity. The center of gravity will now be offset from its previous position, but it will be offset by the same amount whether the weight is at the 50-lb mark or at the zero mark. Its effect in either case is to change the torque applied by the sliding weight to the beam by $wd$, where $w$ is the weight and $d$ is the distance by which the center of gravity has been shifted. Re-zeroing the scale cancels the effect of this change in applied torque by applying an equal torque in the opposite direction. This cancelling torque also is the same whether the sliding weight is at the zero mark or at the 50-lb mark. So, shifting the center of gravity has no net effect if it's followed by re-zeroing of the scale.

Does that make sense?

 

[jack action]

I did some calculations and it actually doesn't matter if your CG is off or not: By resetting your zero with a new weight, you will always be off.

Say your lever arm is 6" and you've calculated 5 g should be added to your 280 g weight. This gives you a needed torque of 1710 g.in to accurately compensate your 50 lb (units are not important).

Now you go back to the zero position. Let's say it was at a 0.075" lever arm length and it was previously zeroed correctly with 280 g (so a torque of 21 g.in is needed to compensate for the scale tray). With your new 285 g weight, the lever arm should be 0.07368" (= 21 / 285) to zero it again. A difference of 0.00132". So when you will go to your original 6" position, the actual lever arm will be 5.99868" (= 6.00000 - 0.00132) giving you a torque of 1709.6238 g.in instead of the 1710 g.in you wanted.

All of this is basically the last two equations of my previous post, defining $L_0$ in both zero situations.

Although, I realize that it is a very small difference (with those numbers; I don't know how realistic they are) and that it doesn't make the scale tip on the other side like you've experienced.

 

[Y.M.E.]

Now you go back to the zero position. Let's say it was at a 0.075" lever arm length and it was previously zeroed correctly with 280 g (so a torque of 21 g.in is needed to compensate for the scale tray). With your new 285 g weight, the lever arm should be 0.07368" (= 21 / 285) to zero it again. A difference of 0.00132". So when you will go to your original 6" position, the actual lever arm will be 5.99868" (= 6.00000 - 0.00132) giving you a torque of 1709.6238 g.in instead of the 1710 g.in you wanted.

I'm not sure I'm following you correctly, but it sounds like you're supposing that re-zeroing the scale is accomplished by changing the position of the zero mark (but not the position of the 50-lb mark). But that's not how the scale works. The zero mark is printed on the beam; it can't move. Re-zeroing is accomplished by moving a separate weight (not the 280-g weight) that is present specifically for the purpose of re-zeroing. The goal of zeroing the scale is that the beam should balance (i.e., be horizontal) when the 280-g weight (or the 285-g weight, as the case may be) is at the unmovable zero mark.

 

[jack action]

Sorry, I thought it was done by changing the fulcrum position. I don't really work with such a scale. But done the way you are explaining it, it does compensate for the CG displacement along the beam.

I guess my only possible explanation left is for the displacement of the CG along the direction perpendicular to the beam.

 

[elmecheng]

I don't think so. Here's why:

You got $m_1$ that is your 51 lb, $m_2$ that is your 280g and their respective lengths from the fulcrum, $L_1$ and $L_2$, such that:
$$m_1 = m_2 \frac{L_2}{L_1}$$
Then you accidentally increase your length by $L_a$ when adding the additional mass $m_a$:
$$m_1 = \left( m_2 + m_a \right) \frac{L_2 + L_a}{L_1}$$
So you decide to zeroing your scale again. Now $m_1$ is replace by $m_0$ (the weight of the scale tray without the 51 lb) and $L_2$ by $L_0$:
$$m_0 = \left( m_2 + m_a \right) \frac{L_0 + L_a}{L_1}$$
You then zero it by increasing the lever arm by $L_x$ (could be negative, if decreased):
$$m_0 = \left( m_2 + m_a \right) \frac{L_0 + L_a + L_x}{L_1}$$
Then you go back to your 51 lb measurement:
$$m_1 = \left( m_2 + m_a \right) \frac{L_2 + L_a + L_x}{L_1}$$
For this to respect your first calculation the following has to be true:
$$\frac{L_2}{L_1} = \frac{L_2 + L_a + L_x}{L_1}$$
Which means that:
$$L_a = - L_x$$
Going back to the zeroing equation:
$$m_0 = \left( m_2 + m_a \right) \frac{L_0 + \left(- L_x\right) + L_x}{L_1} \\ m_0 = \left( m_2 + m_a \right) \frac{L_0}{L_1}$$
Which is impossible since we already know that:
$$m_0 = m_2 \frac{L_0}{L_1}$$

But I did this quickly, maybe I overlooked something.

You're equations are right from a theoretical standpoint. I read that expected sharp edges appear to never have been sharp. Therefore your equations miss a tolerance for each L. I came across a similar issue when I designed a simple scale for a class. The linkages fit but we never built in adjustments to compensate. A manufacturing error outside of tolerance can cause large percentage errors in conversion with mechanical calculators such as this.

I would organize your problem and list your constants such as your scale markings since they can't change. Use the equations for balance that was posted. If you know how to calculate mechanical linkages that would help too in case there is a defect. This seems tedious but doing math is a lot easier compared to shaving off weight and trying the scale again.