this is the situation; I want to build a simple balance to weight up to 5 or 10 grams, I´ve came up with this design which consists of three arms; the one to the left is the weighting arm, where object to be weighted will be hung on a free swiveling hook, so the force is always applied on the same point; the fixed mass arm, the same lenght as the first one, where a fixed mass (either 5 or 10 grams) will be added plus a pointer to errr… well, point in a scale. Finally a third counterweight arm that will balance the other 2 arms minus the fixed weight so the net operating force in the system will be the fixed mass and the mass to be measured.
The problem starts here and I´ve had an hour and half lenght discusion with my father about it… what kind of scale should be used. My opinion is that THIS one is the correct, since the marks are not evenly spaced along the arc but along the height of the scale. But my father says that a scale like THIS should be used, with the marks evenly spaced. The measurements are just an example, what matters is the spacing of the marks.
My scale is based in some obscures memories of vectors and stuff from highschool physics and my father threw out a lot of mumbo-jumbo of radians and angular this and yadda-yadda that, which I really didn´t quite catch. 
So, in the end who´ll be entitled to say “I told you so”, the father or the son? 
I think I used to have a scale a lot like that one, and the scale was something like your version. As you kept pulling down on the hook, it would get harder to move the pointer. In other words, the same amount of force moves the pointer a shorter distance at the top of the scale than at the bottom, so your scale would be correct. I’m certainly not 100% sure about this, though, so don’t start “I told you so”-ing just yet. However, I do think it’s possible that your father is trying to make things harder than they really are.
You are more correct, but neither of you got the zero point right. Think about where the pointer is going to point when there is no mass being measured. (Hint: it’s not straight down).
Aside from the fact that the counterbalance will tip over the top before the pointer weight gets near the horizontal (throwing the whole thing into an entirely diffent conficguration), the mass being measured will move the pointer arm over less of an arc distance for an equal increase in the measured mass as the pointer arm approaches the horizontal.
John Mace makes a valid observation about the counterweight; in fact it may not be necessary at all, I´d just have to move the zero point.
Why make this more complicated than a straight arm bisected in the middle with a counterweight on one end and a (non-weighted) pointer arm pointing straight down? Or, better yet, just the conventional balance scale design where the the two arms form an inverted “V”?
Well, there are some design considerations into this; first, the thing has to be relatively compact, and the rather long arc of movement of the design gives (or should give) better accuracity. I´ll need to measure in 0.05 grams increments. Also it has to be simple and lightweight, yet strong.
To talk straight I´ll use it to work on small indoor flying planes; the last one I made weights 400 miligrams, and I´d like to have a way to measure the different components to see where and how to save weight. The balance should fit into my field box, which is quite small; and of course I´ll be carrying it around so it has to be light and strong.
And of course, I´m prone to come up with odd designs… I have to mantain my reputation.
Still, why introduce all the odd angles? Just adjust the moment arm and/or the counterweight.
I’m assuming by what you say that the counterweight is sized such that, with no mass on the hook, the fixed mass points straight down. In other words, the two arms + counterweight is balanced around the axis.
In this case, as you add mass to the hook, the CG of the two masses (hook mass and fixed mass) will move along a line connecting the fixed mass and the hook. The CG will always be pointing down. It should be intuitive that when your hook mass is equal to your fixed mass, the CG is halfway between. In general, the position of the CG is L/(1+e), where L is the distance between the masses, and e is the weight of the hooked mass (as a fraction of the fixed mass). Note the shape of this function as e gets large.
Also, when the hook mass is very large, the hook will be pointing almost but not quite, straight down. Since both your and your father’s scale tops out at some number (18, or 20, or whatever), it should be obvious that neither of you are right. You can calculate the correct scale spacing given the starting lengths, weights, and angles of your setup using some geometry.
I¢¥m still trying to figure this out; I¢¥ll guess I¢¥ll have to go back to the highschool physics book and see what kind of forces diagram applie to this case. The whole point is that I want to make a balance that uses a fixed weight as reference, and a scale derived from that weight; for that I have to make an appropriate scale, like in example, for X weight measured the fixed mass arm will rotate Y degrees, the problem comes for 2X weight, how many more degrees will the arm rotate, 2Y? log2(Y)? for a WAG example, what gives?, or should I make a vector diagram for every measuer 1X, 2X, 3X, etc?
I thing that trigonometry comes into it… but it¢¥s been too long since highschool and I¢¥m a bit confused.
Let¢¥s put this like a physics problem.
A is the hook point
B is the fixed mass point
C is the pivot point
W is the fixed weight
F is the force applied to the hook (weight to be measured)
And ¥á the angle of rotation of the arms when a force F is applied, respective to the vertical line (the rest position of the CB arm)
Let¢¥s suppose that the arms have zero weight for simplicity, so we don`t need a counterweight to cancell any differences. Also no friction at the pivot.
So, we have the AC segment = CB = 10 cm
W = 5 grams
If F = 1 gram, then ¥á = ?
And so forth for any F…
I don`t know what equation could be used to solve this… :dubious:
Zut, indeed the hook arm would never reach perfect verticality, no matter how much weight you can realistically add; but it¢¥s not supposed to do that any way, when the pointer reaches the scale top the hook arm will lay at an angle of 45¨¬.
Whoa!!! that is some messed up post up there… :eek:
I will try to correct the important part:
"Let`s put this like a physics problem.
A is the hook point
B is the fixed mass point
C is the pivot point
W is the fixed weight
F is the force applied to the hook (weight to be measured)
And ALPHA the angle of rotation of the arms when a force F is applied, respective to the vertical line (the rest position of the CB arm)
Lets suppose that the arms have zero weight for simplicity, so we dont need a counterweight to cancell any differences. Also no friction at the pivot.
So, we have the AC segment = CB = 10 cm
W = 5 grams
If F = 1 gram, then ALPHA = ?
And so forth for any F…
I don`t know what equation could be used to solve this… "
Sure, but my basic point was that the scale represents a function that associates degrees with weight. However, the two functions that you propose (your’s and your dad’s) top out at some number, they can’t be the correct function, because the correct function would not top out, and large weights coud register (as you intuitively realize).
One thing you didn’t state was what the angle is between AC and CB. I infer that the angle is 135 degrees. So here’s how you can do the calculations:
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Realize that the “down” direction will be toward the CM of the two weights. The CM of the two weights falls along the line AB (which has a length L of 9.239, for reference)
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The CM is the point where the weights W and F balance; i.e., if x is the distance between B and the CM, then xW = (L-x)F, or x = (LF)/(W+F). For W=5 and F=1, x= 9.239*1/6= 1.540.
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Now you have the triangle B-C-CM, where one side is x, and one side is 10. The angle at C is the angle which the scale turns (alpha, in other words). You can do whatever triangle geometry is easiest to calculate alpha.
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For example, choose the midpoint of AB; call that point D. The segment CD has a length of 3.827. From the triangle D-C-CM, the angle at C is asin (L/2-x)/3.827 = 53.58 degrees for the 1g weight. Finally, alpha = 135/2-this angle, which is 13.91 degrees for the 1g weight. So 1g maps to 13.91 degrees. And so forth.
Let’s simplify the problem a bit:
We have two arms of lengths n and m at right angles, their weight is negligible and they pivot around the junction. There is a weigth W at the end of M. With no weight on the other arm M is vertical. Now we hang a weight P on the other arm n and the warms rotate an angle a. We want to express P as a function of W, n, m and a.
P * n * cos (a) = W * m * sin (a)
so
P = W * (m/n) * tan(a)
Or find a set of calibrated weights, and weigh them and mark the appropriate places on your scale.
I once had a Calculus teacher who declined to figure such things for real life situations. There can be factors that you don’t enter into the equations, can’t measure (practically,) or that you aren’t aware of that can throw you calcluations off.
A good example came when we were working with calculating the volumes of different volumes made of intersecting shapes. The instructor told us of having been asked to figure the marks for a rod to read the volume of gasoline in a buried tank. Calcluating the marks would be pretty trivial. It wouldn’t have done the tank owner any good, though. If the tank were not perfectly level, or had a dent in it somewhere, or if the measurements were just a little off on the size of the tank, then the rod would be inaccurate. The teacher told the guy to do it the easy way. Get a tank truck with an accurate meter and pump the tank full - and stop and dip the rod in and mark it for certain volumes.
Given the potential for errors (friction in the pivot, mass of the weight and counterweight, necessity of accounting for the effects of the mass of the arms at different angles, possible flex in the arms, etc.) I would personally prefer to borrow a set of accurate weights and weigh them and mark my scale.
Zut, finding the CM of the system proved to be an easy and elegant way to solve the problem Thanks a bunch, it actually seems to work pretty good.
But I don`t quite understand how you did this:
"From the triangle D-C-CM, the angle at C is asin (L/2-x)/3.827 = 53.58 degrees for the 1g weight. " Asin is supposed to be sin?, in any case I get a completely different result… not to worry thou, because I´ll work out that myself, the cogs are turning up there in the brain-box and I´m starting to remember highschool trigonometry.
Mort Furd; I´ve already made a simple balance with the method you mention, using weights to mark the pointer position… works, but it´s messy so to speak, the marks end up being rather irregular. As for the sources of potential errors I think I can cope with them; to begin with I´ll make the pivot point using a clock´s escapemen anchor flywheel, one end of the axis will be mounted on the original ruby bearing and the other will float on a magnetic field (it isn´t as strange as it sounds); so pivot friction will be of no concern. 
The flywheel has three arms at 120º from each other, each for the fixed mass weight, the weighting arm and the counterweight. The nearly zero friction of the pivot will allow me to balance the arms (minus fixed mass) so the center of mass will be at the pivot, thus there won`t be any problem at different angles; and last I wouldn´t care much about flexing of the arms; as I said I´ll use this scale to weight masses of less than 5/10 grams; not enough to flex in any meaningful way the arms.
And of course; beign a Do-It-Yourself kind of guy and mechanical gizmo geek I´d rather make it my way.
Asin = arcsine. However, looking at it again, I realized I made a goof: this should be arctangent (L/2-x is one leg of the triangle, and 3.827 is another).