## How Much Does a Curling Stone Weigh?

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## How Much Do Stones Weigh?

This box is about 10 1/8 inches long, 6 1/4 inches wide and about 4 1/4 inches high. It holds about 16 2/3 pounds of mixed tumbled stones.

#### How Many Pounds of Stone Will I Need?

How many pounds of tumbled stones will you need to use them as ground cover in a potted plant or to fill a decorative jar?

It is very easy to estimate the amount of stone needed if you know the size of the jar and how much the stones weigh in pounds per cubic inch.

This works for small projects such as filling a planter, but it can also be used for estimating how many tons of stone will be needed in a landscaping project.

About ten pounds of mixed tumbled stones will be needed to cover the soil surface in each of these planters to a depth of about two inches. Image © Randy Look, iStockphoto.

#### Planter: An Easy Example…

Let’s say that you have some nice planters like the three in the image at right. Each planter has a soil surface area that is nine inches long and nine inches wide. You want to cover the soil surface about two inches deep with mixed tumbled stones. How many pounds will you need?

Your first job is to calculate how many cubic inches of stone will be needed. This is done by multiplying length x width x depth.

**9″ x 9″ x 2″ = 162 inches ^{3}**

Great! We need 162 cubic inches of stone. How many pounds is that?

Mixed tumbled stones weigh about 0.0618 pounds per cubic inch (or about 107 pounds per cubic foot), and we need 162 cubic inches, so we simply multiply…

**162 inches ^{3} x 0.0619 pounds/inch^{3} = 10.01 pounds**

There’s our answer. We need about 10 pounds of mixed tumbled stones for each planter. This number is an estimate and it should be fairly close. However, depending upon the size of the stones and the shape and the size of the container in which they will be used, that number could be a little high or a little low. But, it is a reasonable estimate.

About 4 2/3 pounds of mixed tumbled stones will be needed to cover the soil in this pot to a depth of about 1 1/2 inches. Image © Skip ODonnell, iStockphoto.

#### Round Flower Pot

The amount of stones needed to cover the surface of a round flower pot can easily be calculated. As an example, how many pounds of stone would be needed to cover the soil surface of the flower pot at right to a depth of about 1 1/2 inches?

Our first task is to calculate the soil surface area. The formula for calculating the area of a circle is A = Pi x r^{2}. Pi is equal to approximately 3.14 and our flower pot has an inside radius of 4 inches (a diameter of 8 inches). So by calculation, the surface area is…

**3.14 x 4 inches x 4 inches = 50.24 inches ^{2}**

If we multiply the surface area by the desired depth of coverage the result will be the volume of stones needed.

**50.24 inches ^{2} x 1.5 inches = 75.36 inches^{3}**

Now to get the weight of the stones needed, we multiply the volume by the density of the stones (0.0619 pounds/inch^{3}).

**75.36 inches ^{3} x 0.0619 pounds/inch^{3} = 4.66 pounds**

So, we will need about 4.66 pounds of stone to cover the soil in our flower pot to a depth of about 1 1/2 inches.

For large projects you can use river stone available at lawn and garden stores. Most rounded river stones weigh about 100 pounds per cubic foot. Image © SpiderPlay, iStockphoto.

#### Much Larger Projects…

This same method of calculation can be applied to landscaping projects where rounded river stone is used. Since these are much larger projects we will use pounds per cubic foot for the weight of the stone. Most rounded river stones weigh about 100 pounds per cubic foot.

Let’s say that we want to have a bed of shrubbery along the front of our home which is 85 feet long. And, we want that bed of shrubbery to be 5 feet wide and covered by a layer of river stone 4 inches deep. First, we calculate how many cubic feet of stone are needed by multiplying the length of the bed by its width and by the depth of the stones…

**85 feet x 5 feet x 0.33 feet = 140.25 feet ^{3}**

Then multiply by volume of stone needed by the density of the stone — which is 100 pounds per cubic foot.

**140.25 feet ^{3} x 100 pounds / foot^{3} = 14,025 pounds**

You can see that we will need about 14,025 pounds of stone or a little over 7 tons.

Happy Tumbling!

#### Article Authored by

Bradley Cole: Bradley is the manager of RockTumbler.com and has authored much of the content on this website. He also does customer support, photography, maintains the website, and consults with customers about rock tumbler repair and maintenance. | |

Hobart M. King: Hobart is the owner of RockTumbler.com and has authored much of the content on this website. He has a PhD in geology and is a GIA graduate gemologist. He also writes most of the content for Geology.com. |

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## How much does cubic foot of stone weight

According to the CRC Handbook of Chemistry and Physics, thedensity of dry air at 20 degrees C at 760 mm of mercury (oneatmosphere of pressure) is 1.204 milligrams per cubiccentimeter. 1 cubic foot = 28,316.8467 cubic centimeters. So, dry air weighs 34,093.48 mg per cu.ft. Which is about 1.2 ounces p…er cu.ft. Observation: I think the above is what you would find if you weigh the air in avacuum (the air is in a container of known weight, and thecontainer is being weighed in a vacuum). The above was probablydetermined with knowledge of physics and mathematics, but the ideais the same. Practically speaking on earth’s surface, at 20 degreesC and one atmosphere, the air would weigh nothing at all; perhapsmore accurately the air ‘weighs’ its own displacement in the airitself. It would be perfectly buoyant. Think about weighing 1 cubicfoot of lake water, with the water completely submerged under thesurface of the lake… quite a different result from weighing it ondry land. Unless you are studying buoyancy, it might be more usefulto use things like atmospheric pressure, or the mass of a sample ofgas and its temperature and pressure within the sample, dependingon your need. The question demonstrates the importance of knowingthe standard conditions (stp) used in research. There is nouniversally established set of standard conditions, so attention isabsolutely necessary. See discussion. Follow-up to observation: If you had a container with a volume of one cubic foot andevacuated it, it would weigh less than if it were full of air atstandard temperature and pressure. You would not need to weigh itin a vacuum, unless its weight in vacuum is what you want. By comparing the weights of the evacuated and the full container,you can determine the weight of the air, essentially, in a vacuum.The weight of the air «in air» remains zero. This seems odd,because we are weighing the container «in air», and the weight goesup when we add the air. I think this does not mean that the airweighs something as it hangs out in the air. [see discussion] I think what is happening is that the observed weight of theperfectly evacuated container at one atmosphere will beless than the weight of the container itself , becauseto find the container’s weight we would need to subtract the weightof the volume of air the whole container is displacing. So puttingmass into the container serves the purpose of bringing us closer tothe weight of the container, and doesn’t demonstrate that the airhas weight while suspended in itself. [see discussion] Any chemist will tell you that the molar volume is 22.4 liters at STP — Standard Temperatureand Pressure (1 atmosphere, 32 F). The mass of a mole is equal tothe molecular weight of the molecule it’s made of, multiplied by 1gram. Air doesn’t have a standard definition as to composition, but it’sgenerally about 78% nitrogen, 21% oxygen, and 1% other stuff. Sincenitrogen has a molecular weight of about 28 and oxygen has amolecular weight of about 32, the molecular weight of air is about28.8 and a mole has a mass of 28.8 grams. Dividing 28.8 grams into 22.400 liters, you get 1.286 grams perliter, which isn’t far from the 1.2 grams per liter quoted above, which is not at STP. Arguing that air «weighs nothing»in ordinary conditions ignores the fact that air may be usedto displace other fluids, and ignores the fact that air’s weightgives air momentum. Or consider the lift of a hot air balloon. If the temperature ofthe air in the balloon is 200F and you are ballooning on a wintermorning when it’s 40 out, how much lift does the balloon produce.You need to use absolute temperatures so adjusting to Rankin scale,the air inside the balloon only weighs 499/659 as much as the airoutside the balloon. That’s about 75% as much — and so the lift ofthat balloon is about 25% of the weight of the air that wouldnormally be in that balloon. Hmmm. A back-of-the-envelopecalculation reveals that 25% of nothing is still nothing — and yethot air balloons DO rise. One might as well assert that you only weigh 85 pounds, becauseyou’re only 85 pounds over the Metropolitan Insurance tables, butyou’ll still have to buy your clothes in the Big & Tall store. (MORE)

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