If you’re a CoCoRaHS observer, you have probably observed occasions where rainwater starts to fill up in the outer cylinder of your CoCoRaHS 4″ Rain Gauge. Like you, I have often wondered, “How much rain have I received if the water in the outer cylinder rises up to the 0.10″ mark or the 0.39″ mark or any other mark on the skinny inner cylinder?” Well, living in the tropical climate of Houston, this question has crossed my mind many times. In fact, in 2018 so far, I have reported ten instances of rainfall in excess of 1″.
From a purely mathematical perspective, we can easily determine a formula using the fact that our rain gauge has one circular cylinder positioned inside another circular cylinder. We can calculate the areas and the volumes of each. I go through the real math below, but here is the expected formula:
Expected formula: y = 8.8x + 1.00
… where “x” is the estimated level of water in the outer cylinder (in hundredths of an inch) as measured against the backdrop of the measurement markings on the skinny inner tube, and “y” is the estimated total precipitation amount received.
CoCoRaHS does a good job of simplifying this in their Frequently Asked Questions: Data Entry FAQ section here. It states:
How do I estimate intense rain with more than 1 inch? You do not need to empty the inner tube and overflow tube every few minutes during a very heavy rain. You can use a ruler to measure the height of water in the outer tube. One inch of rain in the overflow tube represents only about 0.90″ of actual precipitation — due to the displacement of water by the inner tube. It is OK to simply estimate the rainfall to the nearest 0.10″ by measuring the water depth using a ruler and then multiplying by 0.9 to account for the displacement. Remember to add the first inch, from the inner tube.
CoCoRaHS states that “one inch of rain in the overflow tube represents only about 0.90″ of actual precipitation.” Since each 0.10″ measurement mark on the inner tube is exactly one inch, plugging that 0.10″ as the “x” in the Expected formula (y = 8.8x + 1.00) would yield a similar result (y = (8.8)(0.10″) + 1.00″ = 1.88″ or close to the 1.90″ estimated by CoCoRaHS.
So, I decided to perform a simple experiment to verify that this expected formula is true, and if not, to derive an empirical formula that can help me approximate the total precipitation received in excess of 1″ before making a more precise measurement. Here is the empirical formula, based solely on the results of the experiment and not on pure algebra:
Empirical formula: y = 8.6x + 1.11
So, if you see that your skinny inner cylinder is full and that water is collecting in your outer cylinder, say for example up to the 0.24″ mark, you can use this formula to make a reasonable estimate of how much rain has collected in your rain gauge. In this example, if water is up to the 0.24″ mark, you would have approximately 3.17″ in total, calculated as y = (8.6)(0.24″) + 1.11 = 3.17″.
Alternatively, if you don’t like using formulas, scan down below and find my “Easy-to-read Diagram” to help you.
Now, on to the details.
I am a volunteer observer and the local county coordinator for Harris County (TX) for the CoCoRaHS network. CoCoRaHS (pronounced “ko-ko-rozz”) is an acronym for the Community Collaborative Rain, Hail and Snow Network, which originated with the Colorado Climate Center at Colorado State University in 1998 thanks in part to the Fort Collins flood a year prior. In the years since, CoCoRaHS now includes thousands of volunteers nationwide and in several international countries who report valuable precipitation observations on a daily basis. If you’d like to learn more about becoming a CoCoRaHS volunteer, please click here.)
Since our observations are used by individuals and organizations both public and private, CoCoRaHS strives to educate its volunteers to maintain a standardized method of obtaining and reporting their daily precipitation observations. This data is then quality-control checked by state coordinators and data monitors for errors in any of the data.
Each time a rain, hail or snow storm crosses an area, volunteers from the area take measurements of the precipitation. These precipitation reports are then recorded on our web site www.cocorahs.org. The data are then displayed and organized for many of our end-users to analyze and apply to daily situations ranging from water resource analysis and severe storm warnings to neighbors comparing how much rain fell in their backyards.
The equipment used – the official CoCoRaHS 4″ Rain Gauge, which is manufactured to US Weather Bureau standards by a company named Productive Alternatives, Inc. – is the most important tool in this reporting activity. Without standardized equipment like this gauge, the data could be contaminated with readings higher or lower than actually received in a particular location.
The rain gauge has three main pieces: a large outer tube, a skinny inner tube with easy-to-read hundredths of an inch measurement markings, and a top-piece funnel (Figure 1).
As can be seen in Figure 1, there is also a plastic mounting piece available to assist in holding the gauge upright on a board.
The 4″ in its name, the CoCoRaHS 4″ Rain Gauge, refers to the diameter of the opening of the top-piece funnel. When put together properly, the top-piece funnel sits comfortably in the top of the skinny inner tube and funnels rainfall into the skinny inner tube. The skinny inner tube holds a full 1″ of water before a slit in the top allows water in excess of 1″ to overflow into the larger outer tube. The larger outer tube can hold an additional 10.30″ of water for a total of approximately 11.30″.
[Note: in this experiment, I observed that when an excess of 1″ of water was poured into the funnel and into the inner tube, there was an additional 0.03″ or so of water that collected in the overlapping space between the funnel and the inner tube. In other words, about 1.03″ of water was held in the inner tube before water spilled out and overflowed into the outer tube.]
Oftentimes, in heavy downpours, I will visually check the gauge to see how much rain has fallen, assuming it is safe from severe weather to do so. It helps with approximating total rainfall received as well as approximating rainfall rates, and it allows me to send significant weather reports to my local National Weather Service office of Houston/Galveston in League City, TX whenever it seems necessary to do so. However, being that Houston is susceptible to heavy downpours in excess of 1″, that skinny inner tube can fill up quickly thus allowing water to overflow into the outer tube. A glance of the water in the outer gauge, against the backdrop of the inner tube and the easy-to-read markings every hundredth of an inch, can give the observer an idea of how much rain has actually fallen. For example, if the water in the outer tube reached up to about the 0.20″ mark of the inner tube, one might guess that a total of 3″ of water is sitting in the gauge (1″ in the inner tube, and an additional 2″ in the outer tube). But, is it really that simple? I used to think so.
MY SCIENCE EXPERIMENT
Being a weather nerd, I wanted to answer the following question: if the skinny inner tube of my rain gauge overflows, how much rain has actually fallen if the additional water in the outer gauge reaches some mark on the inner tube?
Here are the steps I took on three separate tests to determine if I could answer that question.
Step 1: Fill the skinny inner tube to the 1″ mark.
Step 2: Pour the contents of the skinny inner tube into a plastic cup.
Step 3: Repeat Steps 1 & 2 ten times so that a total of ten plastic cups contain 1″ of water from the skinny inner tube.
Step 4: Assemble the rain gauge.
Step 5: Pour contents of one plastic cup into funnel on top of rain gauge.
Step 6: Measure the water in the rain gauge and note the results.
Step 7: Repeat Steps 5 & 6 until all ten plastic cups have been emptied.
In Step 6 of the experiment (see above), I wrote down the level of the water in the outer cylinder against the measurement markings of the inner tube. The results of the three tests, based on eye-balling the level, were as follows (Figure 2):
Graphing the results in the table above yielded the following scatterplot (Figure 3):
If we add a trendline to the scatterplot data in Figure 3, what is the line telling us? It tells us that we can determine, with some level of accuracy, how much actual rain is in our gauge at a quick glance before we take a more precise and official measurement. Using the formula shown in the graphic above, y=8.5934x + 1.1084 or, more simply, y=8.6x + 1.11 , we can solve for the amount of rainfall received over 1 inch if we can see the height of the water as measured against the backdrop of the skinny inner tube.
- Calculated formula: y = 8.6x + 1.11
“x” is the height of the water in the outer cylinder, to the nearest hundredth of an inch, as measured against the backdrop of the skinny inner tube; and
- “y” is the estimated, calculated amount of rain (in inches) received
Using the calculated formula to solve for the amount of water we need to see in the outer tube (“x”) in order to have an amount of rainfall in whole number inches (“y”), we get the following results for “x” (Figure 4):
Note: amounts (y) may not equal exactly whole numbers in inches because of rounding.
If we take these results and add them to a picture of an actual 4″ rain gauge (using annotations), we have a quick-and-dirty way of determining how many inches of rain has collected in our CoCoRaHS 4″ Rain Gauge (Figure 5).
ACTUAL RESULTS VARY FROM EXPECTED RESULTS
From a math perspective, though, the annotated numbers in Figure 5 should be slightly different. Based on volume calculations (as I will discuss below), the water level should rise every 1.136″ (or 0.1136″ on the inner tube markings). However, in practice, every time I poured a new cup of water with exactly 1″ of volume, not all of the water would spill over into the outer tube. There was always approximately 0.03″ of excess water above the 1.00″ mark on the inner tube. This excess water, which should have flowed out through the slit at the top of the inner tube, seems to be a result of the surface tension of the water in contact with the sides of the inner tube and the funnel, which was slightly overlapped inside the top of the inner tube. This is the main reason that my experimental results differ from the expected amounts, which I discuss now.
THE REAL MATH
To understand why the annotated markings in Figure 5 appear to increase every 0.11″ or 0.12″, we turn to mathematics. Here are some specifications that we need to keep in mind:
- Outer Tube
- Inside Diameter: 4.00″
- Outside Diameter: 4.25″
- Height: 12.00″
- Inner Tube
- Inside Diameter: 1.25″
- Outside Diameter 1.50″
- Height (to overflow cut-out): 10.00″
- Height (overall): 10.3125″
Since we have two cylinders whose bottom is in the shape of a circle, we look back at algebraic formulas for the Area of a circle and the Volume of a Cylinder.
- Area of a Circle = (Pi) x (radius)²
- Volume of a Cylinder = Area x height
Given that the skinny inner cylinder is sitting in the middle of the outer cylinder, thereby displacing water in the outer cylinder, we must find the area of the outer tube and subtract the area of the inner tube. That will give us the area and, ultimately, the volume that the water will occupy (Figure 6).
- Area of outer tube = (Pi) x (inside radius)² = 3.14 x (2.00″)² = 12.56 in²
- Area of inner tube = (Pi) x (outside radius)² = 3.14 x (0.75″)² = 1.76625 in²
- Area of donut-shaped figure = Area of shaded area in Figure 6 =
Area of outer tube – Area of inner tube = 12.56 in² – 1.76625 in² = 10.79375 in²
We also need to calculate the volume of 1″ of water that the inner cylinder can hold. When we add an additional 1″ of water to the already-full inner tube, water will spill into and collect in the outer tube, so that volume will be a key metric. In order to get this amount, we must use the inside diameter of the inner tube to help in this calculation. Note that the inner tube is exactly 10″ high before a slit in the top of the tube allows water to spill over.
- Volume of inner cylinder with 1 inch of water = Area of inner tube (using inside radius) x height = (Pi) x (inside radius)² x height = 3.14 x (0.625 in)² x 10 inches =12.265625 in³
So, the volume of 1 inch of water collected in the inner tube (a tube full of water to the 1.00″ marker) is 12.265625 in². If we were to pour that 1″ into the outer cylinder and replace the inner cylinder, thereby displacing the water, to what height will the volume of water rise in the outer cylinder against the backdrop of the measurement markings on the inner cylinder? Let’s find out.
- Volume of inner tube full of 1 inch of water = 12.265625 in³
- Volume of donut-shaped figure = Area of donut-shaped figure (shaded area in Figure 6) multiplied by the height-to-be-calculated (in inches) = (10.79375 in²)(h)
The volume of the inner tube full of 1″ of water should equal the volume of the donut-shaped figure.
- Volume of inner tube full of 1″ of water = Volume of donut-shaped figure
- 12.265625 in³ = (10.79375 in²)(h)
Solving for “h” in the above formula yields the following result:
- h = (12.265625 in³)/(10.79375 in²) = 1.13636 inches
This tells us that the height “h” in the outer cylinder of an additional 1 inch of water should rise 1.13636 inches against the backdrop of the inner tube. Since we have measurement markings on the inner tube, this would translate to the level 0.113636 of the inner tube. In fact, if we continued this reasoning with additional 1 inch of water into the funnel and into the inner tube and spilling over into the outer tube, we would find that each additional inch of water would raise the level of the collected water in the outer cylinder exactly 0.113636 inches.
Therefore, the expected readings, rounded to the nearest hundredth of an inch, should occur at the following measurement markings on the inner cylinder:
- 2″ —> 0.11″
- 3″ —> 0.23″
- 4″ —> 0.34″
- 5″ —> 0.45″
- 6″ —> 0.57″
- 7″ —> 0.68″
- 8″ —> 0.80″
- 9″ —> 0.91″
- 10″ —> 1.02 in
Using these expected readings, the formula should be the following:
- Expected formula: y = 8.8x + 1.0
Again, though, these expected figures differ from the empirical figures I found to be true in three separate tests, a direct result from the excess water collected in the top of the inner tube.
The information presented above is meant to assist me, as a CoCoRaHS observer, when seeking a reasonable estimate only of the total rainfall my rain gauge has collected. The information above is not meant to replace the procedures that CoCoRaHS observers already follow in making daily readings. CoCoRaHS observers should continue to take precise measurements and submit those measurement when necessary, which typically occurs at their daily observation time of 7am every morning.
Also note that the information presented herein should be used as a guide only and is helpful only when there is water collecting in the outer cylinder, indicating greater than 1″ of rainfall has collected in the skinny inner tube and is beginning to overflow into the outer cylinder.
In addition, my results and the calculated formula (based on my data) works well when the level of water in the outer cylinder reaches approximately 0.10″ or higher. I did not attempt to verify this formula for lesser amounts collected in the outer cylinder. But, from a math perspective, the formula should work.
As you noticed, I used both the word “inches” and the double-prime or double-quotation-mark (“) to denote inches. I don’t like being inconsistent like that, but there were times that I felt it necessary to do so.
I also refer to the inner tube as the inner cylinder and the outer tube as the outer cylinder. For purposes of this article, I use cylinder and tube interchangeably.
I encourage other CoCoRaHS observers to review this data, perform this same experiment, and let me know your results.