Precision Balancer
It is not difficult to make a precision balancer yourself.
You only need a meter of aluminium profile strip and some bolts and nuts,
and you have to work very precise. No bearings, just bolts with a part without thread,
directly moving on the aluminium. With real bearings it could be even more accurate,
but this seems not necessary.
The aluminium strip should not bend under the weight of the objects to be measured,
as that will decrease the precision. I used a square U profile of 2mm alu.
The figure shows the essential parameters of the balance: the radius r, and
the height h between the central rotation point and the rotation points where
the masses are hanging.
The smaller h and the larger r, the more precise your balance.
The balance is stable because of two reasons:
1. because of height h, the effective center of mass of the left and right mass
is always below the central rotation point.
2. Friction in the central rotation point of the balance will make it extra stable.
To get an impression of the accuracy of our balance, we want to know
how much the balance will deflect when we put masses under it.
We can easy find the deflection angle, as follows.
We calculate the effective center of mass x
of the two masses hanging under our balance:
x= r (mr - ml) / (ml + mr)
Because this effective center of mass will be straight under the central rotation point
of the balance, we get for the rotation angle b of
the balance when it is stabilised:
b=arctan(x / h).
Filling in the previous equation and setting mr=ml+
Dm, we get:
b=
arctan(r Dm / [
h (2ml+Dm )] ).
Assuming Dm << ml, this simplifies to:
b=
arctan(r Dm / [2h ml] ).
Assuming small deflections (up to say 20 ° out of balance), we have that
arctan(y) ~ y, so we can further simplify to:
b=
r Dm / (2h ml)
Note that the outcome is in radians, multiply by 180/p
to get it in degrees.
Let's assume that we have built a balance with r= 50cm = 0.5m, and
h= 4mm = 0.004m.
What does this mean for our blade balancing job? Here we have that each blade is
to a very good approximation 100g = 0.1kg.
Filling this in in our formula gives us:
b=
0.5*Dm / (2 * 0.004*0.1) =
625 Dm
So if the mass difference is 1 gram, the scales will deflect
625*0.001=0.625 rad = 35°! That is very easy to see. In this range
the scale behaves linearly, so 0.1 gram difference would result in
3.5° deflection.
What happens if we make small errors in the real version of our balance?
Small errors may have large consequences!
If the central rotation point is not exactly in the middle, one side
of the balance itself will weigh heavier than the other side. This is not
a real problem, because we can just add some extra mass to the other side
to get it balanced again.
The height h
of the mass rotation points is not very critical for the accuracy:
if h on one side is slightly larger than on the other, it does not
really affect the balance, only the h in the center counts and can
be calculated by making a line between the mass rotation points and finding
the perpendicular distance to the central rotation point.
But what is critical is that the distance r is the same for both
sides. So if you have drilled the central rotation point 0.3mm to the right, both mass rotation points
should also move exactly 0.3mm to the right. Failure to do this will result in
a balance that balances unequal masses. So how accurate do we have to be?
Assume we have accidentally drilled the hole for the right mass
Dr too close to the middle. If we now would
hang equal masses under our balance the common center of mass would
be Dr/2 to the left. This will cause a
deflection of g=
arctan(Dr/(2 h))
(while it should be 0!).
To get an impression: Using the setup from the previous example, an error
of 1mm would result in an erroneous deflection of 0.46 rad= 27 °!
The naive user would then think the masses to be off by 27/35 gram!
So, if we really want 0.1 g accuracy we should have
Dr
in the order of 0.1mm. I think this with normal tools an accuracy of 0.25mm is reachable,
giving some 0.25g accuracy.
So the trick is to get the left and right r exactly equal.
A way to check this is to hang nearly-equal weights at both sides, notice the deflection,
and then swap both weights to the other side. The deflection should be equal, but now
to the other side. Or, if we have two exactly the same weights, the deflection should
not change at all.
Now this gives us the clue to get it perfect. If we make r on one side
adjustable, we can adjust until the above test shows perfect balance.
An easy way to get adjustable r is to use a U-profile and keep the open
side of the U at the bottom. Then an axis can be made from the left to right side of the
U. If the axis runs slightly slanted (that is, it crosses the U-bar on one and the other side
at different distances), the distance is adjustable by moving the rotation joint over
this axis. See photo.
| the bolt is mounted slightly slanted, so that moving the rotation point to left or right slightly changes the distance to the central rotation point. |
In order to reach high accuracy the central rotation point has to be vibrated a little
during balancing, to overcome frictional forces there. Without vibrating,
the error often will be ±6°, but with vibrating I estimate it at ±2 °.
To check the performance, I put my 100g blades under the balance.
An extra mass of 0.1g causes a deflection of 4° (±2 °) ;
and an extra mass of 0.6g approximately 20° (±2 °).
Predicted was 3.5° and 21° so
that corresponds very good to the actual measurements.
As an extra rough check I balanced two AA batteries of 25g. The deflection was
10 to 15 degrees when the difference was 0.1g (here the error
seems to be larger, probably because the friction is relatively large
with smaller weights). This again fits nicely with the
theoretical model.
Concluding, the balance indeed shows as accurate as calculated.
Balancing down to the decigram is indeed possible with a $5 balance!
© W.Pasman, 16/4/1