# How to weigh a small item without a scale?

I need to weigh a small item. Let's assume that it's up to 20 pounds (~9 kg).

A tabletop scale would make this really easy,

but unfortunately none is available.

Given other household items, is there a way to do a quick measurement faster than running out to a store and buying a new scale?

First ideas:

1. My first thought's to place the item in a small bucket, put it in a sink/tub, then measure the approximation water displacement to back-calculate the weight. But that'd seem to waste a lot of water.

2. Basically the same as (1), but using two cups or bottles of different sizes instead of a bucket and tub/sink. Could extend this to handle high-volume items by balancing the item to be weighed in the center container to use it as a plunger, but would still seem to need an outer container that has sufficient volume for heavier items.

3. Maybe using something like floss to do some sort of pulley-based system against a known weight?

Create a makeshift see-saw out of a board with a fulcrum in the middle to raise it. Put a bucket on each end of it. Move the board on the fulcrum until it balances.

Put the item in one bucket.

In the other bucket, pour water slowly until your see-saw balances.

Measure exactly how many cups of water were in the bucket, and multiply that 0.48 to convert that number to pounds. :)

• I used this once but it was fairly inaccurate Apr 13 '18 at 20:08
• Hmm. Well, it was just an idea. Apr 14 '18 at 0:14
• I had this same idea but instead of water using items of known weight, like packaged food. And this is much easier in metric if using water. Apr 15 '18 at 13:02

This is based on BrettFromLA's seesaw idea.

1. Get a wooden board (any material will do - it just has to be straight).

2. Put 1L of water (1kg) at one end and your item at the other.

3. Carefully put a pencil or other small, fairly thin item under the board to make a see-saw. The see-saw will tilt one way.

4. Move the pencil from one end to the other until the see-saw tilts the other way.

5. Measure with a ruler (the "^" is the pencil; the vertical bars are just to show the distances to measure).

`(Water) (Item)`
`---------------------------(Plank)-----------------------------`
`| ^ |`
`|<----------a------------>|<-------------------b------------->|`

1. The item is then approximately (b/a) kg.

E.g. if the pencil is about 30cm from the water end and about 150cm from the item end, the item is about 5kg.

Naturally, everything is pretty rough, but this should give you a fairly good idea of how much your item weighs.

• I’ve just noticed that the ‘picture’ doesn’t render well on my phone. ‘a’ is the distance between the water and the pencil, and ‘b’ is the distance between the pencil and the item. Apr 19 '18 at 8:50

Do you have a bathroom scale? They're not that accurate for weighing small amounts, but if you weigh yourself, then weigh yourself holding the object, you'll get a reasonable approximation of the weight by subtracting the first weight from the second.

# The volume of the submerged part of a floating object (in cm3) is equal to the weight of the object in grams.

Similar to Stan's answer, but with a shortcut (no need to measure the water), and possibly no water wastage.

# For buoyant, rectangular objects:

I usually need to weigh something because I'm going to ship it, which means it will be packed in a rectangular box, which almost always has enough air inside for buoyancy.

I'm going to assume we're weighing such a box.

## Steps

1. Find or create a body of water big enough for your box to float in. I used a big cooking pot. If you don't want to waste water, use a pond, or leave the drain closed while you shower.
2. Unless it's waterproof, put the box in a plastic bag. Leave the top of the bag open so the air between the box and the bag can escape.
3. Lower the bag and box into the water until it is floating.
4. Mark the water level on the box inside the bag (the top of the box should be level).
5. Remove the box and bag from the water.
6. Measure the length and width of the box, and the height up to the water line you marked, in centimeters.
7. Multiply these three measurements. This gives you the volume of the submerged part of the box in cm3, which is the volume of the water displaced.
8. Since 1 cm3 of water is 1 g, this number is also the weight of the water displaced in grams.
9. For buoyant objects, the weight of the water displaced is equal to the weight of the object, so the volume of the submerged part of the object in cm3 is equal to the weight of the object in grams.

# For irregularly-shaped objects and non-buoyant objects:

Put the object in a rectangular box that is big enough to make the object float (the technical term for such a box is "a boat"). Use the steps above to get the weight of the box by itself `A`, and the weight of the box with the object `A + B`. Then subtract to get the weight of the object `(A + B) - A = B`.

Making your own scale out of a floating box with tic marks down the side, where you read the weight off the water level, is left as an exercise for the reader.

You're right. Go with your first idea. It's the most accurate and fastest way

You don't need a scale if you can measure the volume of water displaced by your package.

One ml of water weighs one gram at sea level at 20°C.

Twenty pounds weighs 9,072 grams equivalent to 9 litres (a litre weighs a Kilogram) + 72 ml. 1. To waste the minimum amount of water and to get the most accurate measure of weight, find an appropriate container that will completely hold your "thing."
2. Fill a basin exactly to the brim with the empty dry container inside.
3. Put your "thing" into the container and lower it carefully into the basin to displace the water in the container.
4. Measure the water volume displaced and convert into desired units.
5. Done.

If you don't want to waste the water used in the exercise, work carefully to catch the overflow to use for whatever you wish and very little will be lost.

• This will only work with buoyant objects. If you do this with a rock, or an equal-sized piece of gold, the same amount of water will be displaced even though the objects weigh different amounts. Apr 12 '18 at 17:18
• @BrettFromLA EUREKA! I've discovered Archimedes.
– Stan
Apr 12 '18 at 17:40
• @Stan If your answer is incorrect (it is), you should delete it, not ask people to down-vote it. Apr 12 '18 at 20:11
• @RobertCartaino and BretFromLA Fixed. Thanx.
– Stan
Apr 12 '18 at 20:37

You can make a scale. This is a basic lever-word problem.

In order to solve it by this means, however, you will need to use some tools. The accuracy of your answer depends on the accuracy of these tools.

1. You must have an object whose weight you know.
2. You must have an accurate ruler.
3. A straight stiff beam to use as a balance beam.
4. A fulcrum that allows the beam to balance freely.

Here's a sketch of the set-up Here's how you solve the problem using a known weight for reference:

1. Multiply known weight by the distance from the fulcrum (d1)
2. Divide that amount by the distance from the fulcrum to the unknown weight (d2).

Thus, Unknown weight = Known weight x d1 / d2

• If you go to any market in the east, you can see street venders weighing your order of spices on a small hand held balance scale such as this. The one I have from China has a 1 Kilo weight with a ring on one end and a sharp hook on the other (to hold chicken feet or bag loops).
– Stan
Jul 22 '19 at 0:07
• Reader, please note that this answer is nearly identical in concept to that given by Lawrence but for being hung rather than balanced on a fulcrum which increases accuracy and sensitivity.
– Stan
Jul 22 '19 at 2:32