Gordon said: our week at PSA
...Paul is running for the exits! :))
@Gordon: finally looked up Freddy's info. Boy was I wrong!
Born Farrokh Bulsara in Zanzibar!
...Paul is running for the exits! :))
@Gordon: finally looked up Freddy's info. Boy was I wrong!
Born Farrokh Bulsara in Zanzibar!
Yes of Iranian descent and grew up in Zanzibar and India.
birddogthecat said: How to calculate the volume of a cone and then a truncated cone.
Fun. I was imagining a trapezoid, rather than a cone. The area of a trapezoid is very easy; volume, just slightly more complicated.
I find geometry to hold a certain magic and adored it as a student.
@elk
Elk, I believe you are referring to a Trapezoidal Prisim a solid figure which has both volume & surface area. In the example that @Gordon posited, he wrote of a tapered mine shaft. When I sketched what I thought this figure would look like, I came up with the Truncated Cone. Of course, a Truncated Cone is an idealized representation of a mine shaft.
Tomorrow in Chapter 2, we will do the calculations of water depth (or height depending upon perspective) given any volume of water poured into Gordon’s mine shaft.
@wglenn Zanzibar? Where the hell is that?
It is a group of islands off of the east coast of Tanzania.
@elk Well that clears that up. Lol
East Africa, about two-thirds of the way south, on the coast. If you think of Africa as looking like a horse’s head, it is where the eye of the horse would be.
(Lake Victoria is the “eye.” The southern half of lake Victoria is in northern Tanzania; northern half, Uganda.)
@Gordon Chapter 2: Solve for Height given a volume of water in a Truncated Cone (or Frustum Cone)
Happy Labor Day! Sorry about the delay for Chapter 2. I have many excuses why my homework is late. Well, one really. I spent the day at the beach with Patricia and then a cook out with friends. Today, is pretty much the same schedule.
Notwithstanding, I am sure most of you have already figured out the solution to Gordon’s mine shaft volume problem considering a great deal of the work was already done in Chapter 1
In Chapter 1 we determined the total volume of the Truncated Cone. Then we went through the laborious [this is Labor Day after all] procedure of dividing that amount by 2 arriving at 400.6 gallons.
So, we now know the following: we have radii r1 & r2, the volume of water 400.6 gallons, and of course the value of Pi. What we do not know is height. Let us solve for “h” (height) using the same equation from Chapter 1 and we are done with today’s lesson. [see attachment]
Enjoy your holiday everyone and let me know your answers!
PS: Tomorrow is Chapter 3: Plug & Chug
we could also fill the shaft (assuming that I am wrong and that it is not a postive grade shaft an won’t leak, people do make pos grade shafts in mountains) with water from a hose with a flow meter til it is full, then drain half, go in, and measure. nothing in his problem formulation said you had to do it from a desk. (still assuming that ‘area’ in the original problem was ‘volume’ otherwise if the floor of the shaft is 14x14 you can get half the area covered with water but, apart from surface tension, which will be variable depending on the floor material’s hydrophilic or -phobic properties, it’ll be pretty flat).
I am thinking the question has something to do with how much red wine Gordon needs to buy to fill himself half full before doing an DIY upgrade video.
@dcastle That is funny. I now have to admit to another assumption. Since we are audiophiles I came up with this equation: Audiophile = Physical Laziness.
To complete David’s solution, we would require shovels, hose, tape measure, incline gauge, and flow meter. PLUS - manual labor.
@birddogthecat - whatever the answer is, I think my solution is coming out tops b/c it involves going to the hardware store.
Yes, Dave’s solution requires Manuel Labor… And José Labor and Guillermo Labor and… I hear that Day Labor can be had cheap - if you can find him. :))
This all still ignores the fact that the original question didn’t ask how much, it asked how HIGH if half of the area were filled. It didn’t give us enough information to answer how much without making some assumptions. Made up data is not scientifically valid.
Besides, this all sounds like work and I happen to be an audiophile - see equation posted above. 
J.P.
@wingsounds13
J.P. - Gordon’s mine shaft problem does give all the information we need to solve for height of the truncated cone when filled with 1/2 volume of water. If you read my post titled Chapter 2, you will quickly see that we are solving for height not volume.
Also, you cannot fill 1/2 an area as area is not a solid object. Area applies to two dimensional figures.
1) Just start by substituting our known values for the variables r1, r2, &V.
2) Now subtract the value of V from both sides of the equation leaving left of the equation sign blank
3) next we isolate the variable “h” by dividing both sides of the equation by "h"
3a) Alternative Solution: 1st divide both sides of the equation by “h” (before subtracting the value of “V”). You will have “V/h” on the left and on the right side no “h” because the 2 “h” variables on the right will cancel each other out. Now multiply both sides of the equation by “1/V” thus isolating “h” height to the left of the = sign. REMEMBER, we know the value of “V”.
4) we now have the variable “h” isolated to the left of the = sign
5) all that is left to do is solve for “h” (height) by applying 9th grade math skills.
I am sure @Gordon & his nephew will have an answer for us tomorrow.
I also took into account occupational health and safety, since I was writing on labor day in particular. Given the shape of the mineshaft, 1/2 volume of water would be less than half the total height, and lower than the average person’s height legally allowed to work under these conditions. The direct measurement technique keeps hard working people in jobs!
birddogthecat said: all that is left to do is solve for "h" (height) by applying 9th grade math skills.
Uh, oh . . .
birddogthecat said: applying 9th grade math skills.
That's above my pay grade. :(|)
@Gordon Below is the solution that I propose to your nephew’s mine shaft [Truncated or Frustum Cone Cone]:
V= 1/3 Pi x h x (r1^2 + r2^2 + r1 x r2)
R1 = 7’
R2 = 5’
h = 7’
Substitution:
V = 1/3Pi x 7’ x (7^2 + 5^2 + 7 x 5)
V = 1/3Pi x 7’ x 109
V = 1.05 x 7 x 109
Total V = 801.15gallons
Solving h (Height) for 1/2 volume:
801.15/2 = 1/3Pi x h x 109
400.6 = 1.05 x h x 109
400.6 = 114.45h
400.6/114.45 = h
3.5’ (rounded) = h
@Gordon - Check my math but I think this is correct.
Now my head is starting to hurt! I used to be good at math, but that got beat out of me back in 10th grade. Damn teacher insisted on having to write out EVERY step in a solution and would not accept my intuitive but correct leaps. 
Took all the fun out of it. 
Maybe I misunderstand your solution, but the half volume height of a trapezoidal space IS NOT half the height of the space. My off the cuff guess is about 2.8’ but I have not done any of the math to back that up.
J.P.
The puzzle is not asking for a three dimensional solution (volume), rather a two dimensional solution (area) so the above exercise is unnecessarily complicated. Only two dimensions (height and width) are defined - the length of the tunnel is not defined.
J.P.