The temperature at the time of my run was 84 degrees according to weather.com. The dew point was 71 degrees and humidity was at 64%. I learned last summer that the closer the dew point to the actual temperature, the worse the run will feel because the dew point is the temperature at which the moisture is evaporated into the air, which effects the humidity and the general feel of the conditions. According to weather.com the real feel was 89 degrees. The bank sign said it was 94 degrees, which is 5 degrees off which makes me doubt weather.com, but is fairly accurate to say that the conditions for running today were not ideal... heat training is good for you if you do it right, and today, I did it right.
I started with a hand held 20 oz bottle of water and ran an out and back loop of 4 miles to test the feel and hydration effort (I left my camelbak with the 50 oz bladder and 10 oz homemade sports drink on the porch awaiting my results). I decided to drink liberally and run conservatively to make sure I didn't dehydrate. By the time I got home, I had drained the 20 oz of water, which is a rate of ... anyone... anyone... 5 oz per mile. That meant that my camelbak with 50 oz of water could sustain a run of... anyone... anyone... 10 miles total- with the emergency 10 oz of homemade sports drink just in case (I TBS chia seed, 1 TBS of maple syrup, a pinch of sea salt, and 10 oz water).
I grabbed the camelbak and continued the run, deciding not to make a distance decision until later. At about mile 6, I decided that each mile felt more like an additional half mile rather than the regular mile, so at mile 6 it felt like I had run... anyone... anyone... 9 miles. I made the conservative decision to not push it today and run for a total of 10 miles, keeping in mind I had run 4 miles, plus 2, so I only needed to run... anyone... anyone... 4 more miles. I have premeasured routes that I follow so mapping out the route was already done.
Keeping in mind that I was consuming at a rate of 5 oz per mile, and I put my 50 oz pack after 4 miles, needing only to run an additional 6 miles with the 50 oz... how much water would I need to sustain the 6 additional miles AND how much would I have left over at the end of the run (ignore the emergency 10 oz homemade sports drink)... anyone... anyone... I would need 30 oz and would have 20 oz left over- plenty to sustain a conservative run while drinking liberally (this is why I don't affiliate with any political parties).
If I ran 10 miles and each mile felt like an additional 1/2 mile, how long was my effort? Anyone... anyone... 15 miles.
Upon returning home, I had more math to do for my cool down. I had just filled my son's new wading pool and he was having a jolly good time playing in it and wanted me to join in the fun... sounded good to me, so I got in.
The diameter of the pool is 48 inches and the fill line is 10 in. So, the area of the pool is... anyone... anyone... well, let's do a quick lesson on the area of a circle.
To calculate the area of a circle you multiply pi (roughly 3.14, pi is the ratio of the circumference of a circle to its diameter and the decimal goes on forever, so we round it to 3.14) multiplied by the radius squared (to the second power)... pi r squared is how we learn this is school. To find the radius, you simply need to divide the diameter by 2... so the radius of the pool is... anyone... anyone... 24 inches. 24 squared is the same as (24 x 24) or 576. So, pi x 576... I'll wait... anyone... anyone... 1808.64 squared inches. The question- can a pool with the area of 1808.64 inches squared hold a grown man who is 5' 11" with a little boy jumping in and splashing him? The answer... yes.
What's the volume of that pool, you ask? Take the area and multiply it by its height. its height is 10 in, so the volume is 1808.64 x 10 = 18086.4 cubic inches... I include this because of the scientific concept of water displacement of a grown man and a little boy in this pool. But mostly because of his next request. Now, he wanted to play pass the ball (with a beachball) in this pool with daddy (me), mommy, and himself.
So the question. Can a pool with an area of 1808.64 in. sq. and a volume of 18086.4 in. cu. hold a grown man who is 5' 11", a grown woman who is 5' 4" and a little boy who is 43" tall? The answer, yes!
Now, you've done a good job today, class, perhaps later, I'll get into the physics of how to place a slip -n- slide on the yard so that all the water-spouts work correctly so that the little boy can have fun slipping AND sliding... that, my friends, was the real adventure of the day.
*just in case you're wondering... yes, this is what I think about when I am running
Enjoy!
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