Mar 30
I am NOT a math guy - so help! :)
on March 30th, 2007 - UncategorizedAdd comments
We received four inches of rain today.
I have an outdoor pool that is 38 feet X 22 feet (I don’t think depth matters for this, does it?)
Anyway, assuming it did rain 4 inches today, and it rained 4 inches equally over the entire pool, how much water (in gallons) did my pool trap today?
No, I have no idea why I wonder this. I just do 
Thanks,
Rob
*** Random Post ***
March 31st, 2007 at 04:20
Two methods:
The magic of metric units:
4 inches = 0.1016 m
38 feet = 11.5824 m
22 feet = 6.7056 m
Multiply…
Volume = 7.89096 m^3
Volume = 2085 gallons
The magic of Google:
http://www.google.com/search?hl=en&q=4+inches+*+22+feet+*+38+feet+in+gallons&btnG=Google+Search
Stu
March 31st, 2007 at 07:14
You should be thoroughly ashamed of yourself.
You don’t need to be a mathematician for this: all you need is just a few years of elementary school.
You are just being mentally and intellectually lazy.
Which is unforgivable.
At LEAST show an attempt (and you will be ashamed of yourself too, once you realize how childishly easy it is), and THEN I MIGHT tell you if you’re right or wrong!
(Oh, and yes depth DOES matter: I’ll leave it to you to explain the obvious ‘why’)
March 31st, 2007 at 08:05
@Paul - you don’t know either, huh?
I did try it myself - I didn’t believe the number (still don’t believe it)
Rob
March 31st, 2007 at 10:40
Of COURSE I know …
1. The depth matters, because if the accumulated dirt and leaves in your pool have reduced the depth to 1 inch, the +3 inch it rained more than the depth would be irrelevant. I will assume a depth of at least 4 inch though.
2. I foot = 12 inches, so the volume of 4 inch rain in your pool = (38*12) * (22*12) * 4 = 481,536 cubic inch.
3. Since there are 231 cubic inches in a gallon, that volume equates to 481,536 / 231 = a little over 2,084.5 gallon.
That wasn’t really THAT difficult, was it?
March 31st, 2007 at 11:32
@Paul - regarding your points (above):
1) An astute reader of my blog would know that my pool has been drained for some time, therefore rendering the depth a non-issue
2)I did the math this way:
38×22 feet = 836 square feet * 12 = 10032 square inches.
10032 sq inches * 4 = 40,128 cubic inches
40,128 / 231 = 173.71 gallons
Finally - It must be THAT difficults, since we got drastically different answers
Rob
March 31st, 2007 at 11:50
1. Since I AM an astute reader of your blog, I KNEW you drained it. Drained pools however, have a habit of filling up with dirt and dead leaves (and God knows what else… dead bodies?). So actual depth IS an issue, but once you have established (or assumed, as I did) that it IS 4 or more inches deep, then and only then, does it become irrelevant.
2. “836 square feet * 12 = 10032 square inches”
Uh NO!
By Jeeves, what school did you attend?
You’re messing up your dimensions!
Which is easy to see when you take ONE square foot! According to your flawed arihmetic:
1 square foot * 12 = 12 square inches. Believe me: 1 square foot = 144 square inches! You’re dealing with SQUARE feet, so you should multiply by 12 squared.
- Okay, I agree it IS difficult then. But just for one of us.
March 31st, 2007 at 11:53
Hey, I admitted I wasn’t a math guy! I had one algebra class in 1975! No geometry, no trig, no calculus (and I know it shows - that’s why I asked for someone else to give me the right answer! At least I know what my weak points are
Heh…
Rob
March 31st, 2007 at 12:00
In fact, if one is as careless with dimensions as you were, one can easily prove that $1 = 1 cent!
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢
(use a calculator to double check the ‘correctness’! Pretty clever, eh?
March 31st, 2007 at 12:11
“No geometry, no trig, no calculus”
None of it needed!
Which I just proved, since I didn’t have any of that either. Heck, I don’t even know the imperial system! (Metric only). So I had to look up inches, feet and gallons!
I know my weaknesses too, but I don’t let them stop me..
March 31st, 2007 at 12:24
Okay, just to FORCE you to use your brain…
Same situation but now your pool’s bottom slopes 20 degrees lengthwise, with the shallowest part being 4 inches deep.
NOW what is the answer?
(You’ve got two minutes for this one…)
March 31st, 2007 at 12:28
This whole thread reminds me of re-writing Trent’s PRISM Benchmark program (I relaunched it as PRISM Benchmark Pro).
I saw numbers with his tool that didn’t make sense to me - so I asked him for the source. He told me it had been lost in a horrible hunting accident. So I re-wrote the application from scratch. About a year later I found the source on Trent’s lab server (so he lied to me, and two years later when he worked for me - I remembered that lie :))
Anyway, I took my souce to Allen and Wes so often to double-check my math that I know they were sick of seeing me. But now that thousands and thousands of people have used Benchmark Pro, I’m glad I took the extra precaution of getting quadruple-checked. THAT is how much faith I have in my math abilities
March 31st, 2007 at 12:29
@Paul - the answer is the same - he slope doesn’t matter in this case because we are measure the volume of water trapped over a given area - the slope doesn’t come into play.
March 31st, 2007 at 12:32
On one hand I’m glad to see you still know when NOT to apply math … just common sense.
On the other hand .. I’d SOOO hoped you would spend the rest of the weekend trying to figure this one out…
March 31st, 2007 at 13:02
About lying about lost source …
There are probably one or two source files out there that I gladly (but wrongly) would deny authorship of.
(There are more instances where I deliberately have removed my name from source code, but that had more to do with contributions by others
And although I despise lying, if I HAD to lie about losing source code, I most DEFINITELY wouldn’t try making people believe that I lost it in “A HUNTING ACCIDENT”!!
A disk crash… sure… but A HUNTING ACCIDENT? Did I read that right?
March 31st, 2007 at 13:04
March 31st, 2007 at 13:07
Well, knowing the person in question, since the claim was so outlandish, I guess it was his way of telling you you weren’t going to get it. Period.
March 31st, 2007 at 14:25
QUIT worrying about this! By now I am sure all the water in Rob’s pool has dried up, anyway!
March 31st, 2007 at 14:34
March 31st, 2007 at 14:37
No! Can’t leave it alone yet.
I STILL need Rob to do a little thinking.
This time with regards to my proof that $1 = 1 cent.
It SHOULD be fairly easy, since I already explained WHAT KIND of error he made with the volume, and as EXAMPLE of the SAME error I presented my proof that $1 = 1 cent.
So here’s the question: Since common sense dictates that $1 is NOT equal to 1 cent, WHERE in the given formula lies the error?
(10 bonus points for this one)
March 31st, 2007 at 14:37
Besides, that post deserves AT LEAST 20 comments!
March 31st, 2007 at 15:19
Oh, and that’s 10 bonus points for ANYONE who gives the first correct answer!
April 1st, 2007 at 19:14
Paul,
When you say that the shallowest part of the pool is 4 inches you are talking about the pool going from 4 degrees deep to the deep in it would seem that you still have the same 38×22 foot catch basin to receive rain water… thus you would have the same amount of water in the bottom of the pool.
David
April 1st, 2007 at 19:40
Of course at a 20 degree decline the deepest part is only around 14.9 feet… Unless I have the 20 degrees on the wrong side then it turns to around 13.86 feet…
(all done with help from my son…
)
April 1st, 2007 at 19:50
@david … re.: comment #21:
“..you would have the same amount of water ..”
Absolutely correct! No bonus points though, since Rob already answered this correctly in comment #11, and I acknowledged his correct solution to my weak attempt at an April-fools-day math problem in comment #12.
April 1st, 2007 at 21:00
@David… re: comment #22.
This comment has me stumped.
For several reasons.
First of all: why that calculation? You already established that the slope plays no role whatsoever.
second: How can you get the slope on the wrong side? I gave the shallowest point, and said it was length-wise.
third: you need your son’s help with this?
fourth: I’m VERY impressed by your son’s trig AND arithmetic (to compensate for the 4 inches!) skills!
So… now… go for the bonus points! Answer comment 18 (19)
April 1st, 2007 at 23:08
Actually, I compensated for the 4 inches. I just needed help on the formula to figure out the sides of a triangle (long time since I did that math). So the formula according to my son was tang(angle) = opposite/adjecent.
(10¢)^2 = ($0.10 * 10^2)^2
April 2nd, 2007 at 08:02
@David .. correct formula (I don’t have a son, so I used wikipedia instead). And if you try to calculate this in Excel (like I did) the formula is =TAN(20*(PI()/180)) (since Excel insists on the angle being in radians, rather than degrees).
As for the ‘money’ formula you gave:
1. It doesn’t tell me what was wrong with mine!
2. It is wrong too!
April 2nd, 2007 at 11:14
@Paul: I am late to this party, but it is wrong because cents and dollars are different units, you cannot square one of then to get the other. Common sense again.
April 2nd, 2007 at 12:39
@Kami .. That they are different units is NOT a (the!) problem in this case, since you can express one unit as a fraction of the other, so, in essence, you only have one unit (dimension). And even when you express the cents in dollars ($0.01) you STILL end up with an incorrect formula.
But you’re right about the other part of your comment: the problem is in the squaring. That stated equality “100c = (10c)^2″ is, in fact NOT an equality. (10c)^2 is NOT 100c. It is 100(c^2). Of course, I don’t know what a square cent is, but that doesn’t mean you can simply ignore it. The human brain is extremely good at ignoring ‘non-sensical’ data and/or replacing it with something it DOES know, but in math, you can’t do that! 10c * 10 = 100c. 10c * 10c = 100(c^2), something different altogether.