# Simplified triple beam scale

What could possibly be simpler than a triple beam scale? (Ohaus Corp.)

A simplified triple beam scale, of course:

1. Just one identically spaced pair of notches on each beam, allowing 2 possible weight positions per beam for a total of 2 × 2 × 2 = 8 possible combinations between all three beams. (No gradually sliding weights. No attachment weights to hang onto the balance arm.)

2. The beams’ weights need not be positive integers. (A helium balloon, for instance, can be a handy negative weight.)

3. The beams’ weights form an arithmetic progression, where the middle weight is simply the average of the other two weights.

4. An additional 1-unit free weight is available for calibration but may also be added to the weighing pan during measurement.

Such a scale does have a perfectly useful application — to measure out 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 or 16 units of sample in a single weighing, . . .

. . . with what selection of weights on the three beams?

(Notes: Two essentially equivalent solutions. No need to be able to measure 0 units of sample. The distance between each arm’s notches equals the pan’s lever arm, so the arms’ weights are as heavy as the amounts of sample they balance. What are called arms’ weights here are also known as poises and riders.)

• Each beam has 8 possible combinations, but on point 3, the middle weight is the average of other two. Which one is the middle weight here? Also did I miss something that.. a trivial solution like 3 beams of 1-8 can measure 1-24? Apr 29, 2020 at 4:25
• I've gotta clarify that. It's 8 combinations of the 3 beams together. Sorry about that, @athin.
– humn
Apr 29, 2020 at 4:28
• I don't understand the 1-unit free weight. What can you do about it? Apr 29, 2020 at 5:14
• Here's the 1-unit free weight's intended action plan, @Culwer Kwan. Before weighing anything else, the scale can be calibrated by adjusting it to horizontal when the 1-unit weight is by itself in the pan and the beams' weights are positioned to the combination meant to balance 1 unit. Then in measuring some amount of sample, the option of adding the 1 unit weight to the pan is equivalent to subtracting 1 unit of weight on the balance arm.
– humn
Apr 29, 2020 at 5:29

Because the description is a bit messy, I'll start with my interpretation:

1. On the right side, we have three different weights (can be negative) each individually switchable between positions 1 and 2, which we'll call "off" and "on" for convenience.
2. We can adjust the zero position (all three beams "off") of the scale, if we need to.
3. The weights on the right side need to be in linear progression
4. We can add a 1 gram weight on the left side when necessary

Were it not for point 3, we would just use (doubled) binary numbers: to make the numbers 0-7 by combining 1, 2, and 4 is trivial, so if we make the beams add 2g, 4g, and 8g, we can make all even numbers between 0 and 14. We then use the "zero adjust knob" to skew this by 2 grams (so that the "all weights off" situation needs 2 grams to balance), and we can make all even numbers from 2 to 16. From there we get to the odd weights by adding the 1g extra weight to the pan when necessary.

Symmetrically, we could also use -2g, -4g, and -8g as the weights, and adjust the zero so that "all beams off" is 16.

But point 3 is, of course, what this puzzle is all about. 2g, 4g, and 8g are not allowed, because 4g is closer to 2g than 8g; it needs to be in the middle. How can we make that work?

The answer seems to lie in

using negabinary numbers instead of binary numbers.

That neatly spaces the weights on the beams:

using -2g, 4g and -8g (or, symmetrically, 2g, -4g and 8g), we can still make eight consecutive even number weights, which we can "zero adjust" to coincide with 2-16.

This is exactly what we need, and -2 is smack in the middle of -8 and 4, so we have the required linear progression too.

To demonstrate that this works, let's pick the option with only one helium balloon. (Balloons are nice, but I hate subtraction.)

First, let's "zero adjust" so that "all beams off" needs 6g to balance. Then, we can measure like this:

 Desired weight | -4g |  2g |  8g | 1g in pan | zero adjust offset
---------------+-----+-----+-----+-----------+-------------------
1g   | -4g |     |     | -1g       | +6g
2g   | -4g |     |     |           | +6g
3g   | -4g | +2g |     | -1g       | +6g
4g   | -4g | +2g |     |           | +6g
5g   |     |     |     | -1g       | +6g
6g   |     |     |     |           | +6g
7g   |     | +2g |     | -1g       | +6g
8g   |     | +2g |     |           | +6g
9g   | -4g |     | +8g | -1g       | +6g
10g   | -4g |     | +8g |           | +6g
11g   | -4g | +2g | +8g | -1g       | +6g
12g   | -4g | +2g | +8g |           | +6g
13g   |     |     | +8g | -1g       | +6g
14g   |     |     | +8g |           | +6g
15g   |     | +2g | +8g | -1g       | +6g
16g   |     | +2g | +8g |           | +6g


• Can the offset be removed by adding offset/3 to each weights? (And they are still an AP!) Apr 29, 2020 at 12:35
• @athin depends on whether $10 + 4 = 16$, I guess.
– Bass
Apr 29, 2020 at 12:52
• @athin , note from poser: That the spaces between notches are all equal to the pan's lever arm is meant to eliminate an infinity of variations such as adding or multiplying this solution's weights by a constant (other than multiplying by -1, as Bass does here).
– humn
Apr 29, 2020 at 14:04
• @humn That's one of the reasons I called the question "a bit messy" :-) I think you could have used "weights or helium balloons that you are allowed to attach only to the right-hand pan of a balance scale" to pose the same puzzle in a way that eliminates the lever arm length complications entirely. My answer is, I now notice, in grams, which should really be "the unit of weight that is required to balance the minimal configuration change on the right side".
– Bass
Apr 29, 2020 at 16:42
• @humn come to think of it, if you are allowed to only manipulate one balance scale pan, you can probably shape the pan's bottom side so that exactly one helium balloon can be held in place below it, which would result in the solution being unique :-) Do you want to repost this puzzle with those modifications in a year, or may I? :-)
– Bass
Apr 29, 2020 at 16:50