Tried to use a bucket for a rain gauge but realized too late that it was not a perfect cylinder. The diameter at the top was 11.25 inches. The bucket was 14.35 inches tall. The diameter at the bottom was 10.25 inches. It held 7.25 inches of rain. What was the actual depth of rain if we set the bottom diameter to 11.25 inches. I need to know because my daughter trusted me to get an accurate rain depth for the storm we had, and I did not have my regular rain gauge out. But there was an empty 5 gal bucket in the yard, hence this question. Thank you so much for any assistance you can provide!
Rain gauge assistance
- Thread starter DanaB
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Dr.Peterson
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Find the volume of water using a formula for volume of a frustum of a cone (taking the diameter at the depth of the water as one base). Then divide that by the area of the opening at the top of the bucket, to find the effective inches.
I found an explanation of how to do this at the bottom of this page: https://scied.ucar.edu/blog/measuring-rainfall-–-it’s-easy-and-difficult-same-time-0 Unfortunately, the volume formula is wrong! It was a good guess that you could use the average of the radii, and might be close enough for a typical bucket, but the actual formula for a frustum of a cone is [MATH]V = \frac{\pi(R^2+rR+r^2)h}{3}[/MATH], where r is the base radius and R is the top radius (at the level of the water).
Another approach would be to pour the water (or the same amount of water, from the same bucket at the same depth) into a perfect cylinder, measure its depth, and then multiply by the ratio of the square of the diameter of the cylinder to the square of the diameter of the bucket.
Rather than leave you to do the calculations (as this clearly is not homework unless you're a great liar), I'll go ahead and do it.
The bucket is filled to 7.25/14.35 = 0.505 of its height, so the diameter at the top of the water is 0.505 times the difference more than the base; that is, 11.25 + 0.505(11.25-10.25) = 11.755 inches. So we have r = 10.25/2 = 5.125, R = 11.755/2 = 5.878, and h = 7.25. This gives a volume of [MATH]V = \frac{\pi(5.878^2+(5.125)(5.878)+5.125^2)7.25}{3} = 690.69\text{ in}^3[/MATH]. The area of the top opening is [MATH]A = \pi(5.625)^2 = 99.4\text{ in}^2[/MATH], so the effective depth of water is [MATH]\frac{690.69}{99.4} = 6.95\text{ in}[/MATH].
(The web page's volume would be 689.366, which is not far off, as I expected, so you can use his formula if you like. You'll get 6.935 inches for the depth ...)
I found an explanation of how to do this at the bottom of this page: https://scied.ucar.edu/blog/measuring-rainfall-–-it’s-easy-and-difficult-same-time-0 Unfortunately, the volume formula is wrong! It was a good guess that you could use the average of the radii, and might be close enough for a typical bucket, but the actual formula for a frustum of a cone is [MATH]V = \frac{\pi(R^2+rR+r^2)h}{3}[/MATH], where r is the base radius and R is the top radius (at the level of the water).
Another approach would be to pour the water (or the same amount of water, from the same bucket at the same depth) into a perfect cylinder, measure its depth, and then multiply by the ratio of the square of the diameter of the cylinder to the square of the diameter of the bucket.
Rather than leave you to do the calculations (as this clearly is not homework unless you're a great liar), I'll go ahead and do it.
The bucket is filled to 7.25/14.35 = 0.505 of its height, so the diameter at the top of the water is 0.505 times the difference more than the base; that is, 11.25 + 0.505(11.25-10.25) = 11.755 inches. So we have r = 10.25/2 = 5.125, R = 11.755/2 = 5.878, and h = 7.25. This gives a volume of [MATH]V = \frac{\pi(5.878^2+(5.125)(5.878)+5.125^2)7.25}{3} = 690.69\text{ in}^3[/MATH]. The area of the top opening is [MATH]A = \pi(5.625)^2 = 99.4\text{ in}^2[/MATH], so the effective depth of water is [MATH]\frac{690.69}{99.4} = 6.95\text{ in}[/MATH].
(The web page's volume would be 689.366, which is not far off, as I expected, so you can use his formula if you like. You'll get 6.935 inches for the depth ...)
That’s incredible, and surprising! I tried to find a way to create another cylinder of desired dimensions to resolve it, but just couldn’t recreate it with adequate accuracy. I actually thought the difference in values would have been noticeably bigger, but I see the math and have no reason to doubt it. I can’t tell you how much I appreciate this. You have saved me in the eyes of my daughter and you can’t put a price on that! Thank you so much!
D
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There is a way to approximate the height.
Diameter of the top of the water level = 10.25 + (11.25-10.25)/14.35 *7.25 =10.75522648
Average diameter of water (column in the bucket) = (10.25+10.75522648)/2 = 10,50261
corrected length of water column = 7.25 * ( 10,50261/11.25)^2 = 6.318699578 =6.32"
should be close enough for "weather.com"
That is closer to what I had estimated the true depth to be. I do not have the mathematical aptitude to discern whose math is correct but I’m happy to have two calculated responses and will use them to my advantage depending on the audience. Thank you so much for your time!
Dr.Peterson
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Comparing the two, I see that I used the wrong base to find the water-level diameter. Here is a redo of my method. (SK's is identical to the page I referred to, and used the right number.)
The bucket is filled to 7.25/14.35 = 0.505 of its height, so the diameter at the top of the water is 0.505 times the difference more than the base; that is, 10.25 + 0.505(11.25-10.25) = 10.755 inches. So we have r = 10.25/2 = 5.125, R = 10.755/2 = 5.378, and h = 7.25. This gives a volume of [MATH]V = \frac{\pi(5.378^2+(5.125)(5.378)+5.125^2)7.25}{3} = 628.21\text{ in}^3[/MATH]. The area of the top opening is [MATH]A = \pi(5.625)^2 = 99.4\text{ in}^2[/MATH], so the effective depth of water is [MATH]\frac{628.21}{99.4} = 6.32\text{ in}[/MATH].
(The web page's volume would be 628.08, which is not far off, as I expected, so you can use his formula if you like. You'll get 6.32 inches for the depth ...)
So, yes, it's 6.32 inches of rain.
I appreciate the update. Once I got the two responses, I started to look at the calculations. I’m no math major but I do have a masters degree in biology and am no stranger to math. It didn’t take long to see that the water surface couldn’t have a diameter greater than the bucket opening. Rather than go any further I just went with the other answer. What I appreciate now is that both answers agree. Thanks for taking the time to own and correct the mistake.