Everyone loves crazy science questions. Will a bullet shot in the air hurt you when it comes back to earth? (It probably would.) “What would happen to the rotation of the earth if everyone in the world jumped in the air at once?” (Not much, even if they were all grouped in an area as tightly as possible.)
We found some of the strangest swimming pool questions on the Internet and asked some scientists, engineers, and experts to answer them for us.
If I filled my swimming pool with Jell-O and jumped in, how far would I sink?
It depends on the flavor. No seriously, it depends on several factors. According to Professor Michael Schaeffer of Northwestern University, it depends on how much gelatin you dissolve into the water while making the Jell-O. Gelatin is what gives Jell-O its semi-solid consistency and that would be critical to whether or not a person would be able to tread the Jell-O or sink to the bottom of the pool. The average Jell-O mixture consists of less than 2% gelatin, not enough to keep a person afloat in a tasty fruit Jell-O bath. But if you were able somehow to boil the water necessary to fill an average pool with Jell-O (which would require about 2,000 gallons of kerosene, by one estimate), you could increase the level of gelatin to, say, 4-5% and produce a more firm Jell-O that could keep a human body from slicing through to the bottom of the pool – at least initially, and from a modest height. As Schaeffer points out, though, it would be quite an undertaking and not advisable.
Is there a difference in the effort required to swim in the shallow end of a pool as opposed to the deep end?
This brings about the science of fluid dynamics. An object’s motion will be hindered more greatly by any turbulence created by factors nearer to that object. As you swim and create waves, the side and bottom of a pool will create a turbulence effect. If you are next to the side of the pool or in a shallow end where your body is near the bottom, you will naturally receive more of a turbulent impact from the surface of the pool. It would be easier to swim laps in the deep end of the pool.
If you filled one swimming pool with golf balls, another with tennis balls, and a third with bowling balls, which would have the most empty space?
The answer to this one may surprise you. According to a principle of physics called “closed packing,” spherical objects will occupy the same space given an infinite amount of space. Thus, a pool the size of the entire universe packed with marbles, tennis balls, golf balls, or bowling balls will have the same amount of “empty space.”
However, on a smaller scale, such as that of the swimming pool in your backyard, there would be some variation based on the configuration of the different types of balls packed in the pool. The smaller the ball, the less empty space, though the difference would be negligible if the packing of the objects was random (just tossing the balls in as opposed to stacking them with some precision).
Thanks to Oscar Corrales for this answer. He’s an engineer living in Edmond, Oklahoma.
Does being in a swimming pool during a storm increase your chances of being struck by lightning?
US National Weather Service Science and Operations Officer Ted Funk tells us that it’s wise to leave a swimming pool during a lightning storm. “A lightning strike can generate 300 million Volts and 10,000 – 200,000 amps, and a body of water can conduct the lightning’s electricity from a strike.”
While there is no evidence that more people are struck or harmed by lightning while in water (most people are on land at any given time), the physics of lightning storms doesn’t make it a good idea to be in your pool during a severe storm.
If all of the water in my swimming pool became a cloud, how large would that cloud be?
For this one we consulted with a meteorologist: David Littlefield of Chicago, Illinois, who has written several booklets on weather and its effects for various agencies.
First we’ll assume a swimming pool that is filled with 20,000 gallons of water. Assuming a water temperature of 75 degrees nd an air temperature of 65 degrees, we can calculate the amount of the water that would evaporate and the corresponding size in vapor. However, we must also have some of the water remain as droplets, otherwise a cloud cannot form. Understand as well that the surrounding air in the sky would also interact with the vapor and water droplets and contribute to the size and consistency of the cloud.
“There are many, many variables, but with a base starting point of 20,000 gallons of water we can make a good guess,” Littlefield says. The answer is quite staggering: a cloud more than 388 million cubic feet in volume. A cloud that large could cover the Oklahoma Sooner Stadium 1,000 times.
If I point a fan at my swimming pool, will it lower the temperature of the water?
The short answer is: not very much at all.
“The air over the pool water warms the water by conduction, which is governed largely by the temperature difference between the air and water, but it also cools the water by accelerating evaporation, and this is governed by the relative humidity. The warming effect increases with lower water temperature, and the cooling effect increases with lower relative humidity. Warming due to transfer of the air’s kinetic energy will be infinitesimal by comparison. Thus, a very cold pool in 95°F air at 100% RH will be warmed while a slightly cool pool in 95°F at 0% RH will be cooled, with the equilibrium point being at 95°F or lower depending on the humidity.”
How many pennies would it take to fill an average swimming pool?
There are better places to store your pennies, ever hear of a piggy bank?
But if you insist on this folly, we figured this out for you. Let’s assume we have a 16-foot wide and 32-foot long pool with a shallow depth of 4 feet and a deep end depth of 10 feet. The total volume of that swimming pool would be 26,812 gallons.
First, we figured out the volume of a penny.
Volume = Pi *r^2*h
(r is the radius, and h is the height)
For a standard penny, r=.5 inch and h = .123 cm
So the volume of a penny = Pi(.5 in.)^2(.123 cm)^2 = 0.6233 cm^3
Then we had to figure out how many times the volume of a penny could fit into the volume of our swimming pool.
1 US gallon = 3 785.4118 cm^3
3785.4116 cm^3 (1 gallon)/0.6233 cm^3(1 penny) = 6073.
That means it takes 6,073 pennies to fill a gallon.
When we multiply that by the 26,812 gallons in our pool, we arrive at 162,829,276 pennies:
(6073)x(26812 gallons)=162829276
That’s $1,628,292 dollars and 76 cents – a pretty good start to a savings account.
How much water do I lose in a week from my swimming pool via evaporation?
On average, a swimming pool will lose about ¼ of an inch per day due to evaporation, or about 1 ¾ inches in a week. This depends on water volume, wind, sunlight, and humidity, so that figure can change slightly. If you think you’re losing too much water from your pool, check it for leaks.