- 04 Nov '10 15:55So I put my above ground pool, 7 meters across, to sleep for the winter, it freezes here, get maybe 100 mm of ice. But not now. It is raining like the dickens here for a few days. The cover is slanting down at 45 degrees, meeting the waterline 500 mm below where the top of the cover is.

It meets the water and flattens out 500 mm in so the flat part of the cover is 6 meters across. Assuming everything atomically smooth, a truncated cone starting at 7 meters across meets a flat which is 6 meters across.

The rain has caused an accumulation of 40 mm exactly. If it was a true cylindrical section you can say there was 40 mm of rain. But the cone shaped section raises the level of the water a bit. What is the true amount of rain?

Also, how deep does the cone peak go down? the water is 1 meter deep and the wall holding the pool together is 1.50 meters high, so does the peak of the cone extend underground if it was a full cone or does it come to focus above the bottom of the pool if it were real? - 05 Nov '10 04:09 / 1 edit

The cone comes to focus 3.5 meters from the top surface of the pool. As for the height im not sure I follow. Are you talking about taking the volume of the right cirrcular cylinder and filling the cone to see the difference in measured height?*Originally posted by sonhouse***So I put my above ground pool, 7 meters across, to sleep for the winter, it freezes here, get maybe 100 mm of ice. But not now. It is raining like the dickens here for a few days. The cover is slanting down at 45 degrees, meeting the waterline 500 mm below where the top of the cover is.**

It meets the water and flattens out 500 mm in so the flat part of t ...[text shortened]... und if it was a full cone or does it come to focus above the bottom of the pool if it were real?

if you are:

hcone = (3*r_cyl*h_cyl)/(r_max^2 + r_max*r_min + r_min^2)

How are you obtaining the "true" height of 40 mm? - 06 Nov '10 06:34 / 1 edit

I was saying if the height of water measured 40 mm in a cylindrical volume, you could state there was 40 mm of rain. If the cone shaped section blocked off part of the cylindrical volume then 40 mm of measured water level would have been created by less than 40 mm of rain. That is the level we are after.*Originally posted by joe shmo***The cone comes to focus 3.5 meters from the top surface of the pool. As for the height im not sure I follow. Are you talking about taking the volume of the right cirrcular cylinder and filling the cone to see the difference in measured height?**

if you are:

hcone = (3*r_cyl*h_cyl)/(r_max^2 + r_max*r_min + r_min^2)

How are you obtaining the "true" height of 40 mm? - 06 Nov '10 13:26 / 3 edits

ok, its the height measured by a cylinder of equal radius your after.*Originally posted by sonhouse***I was saying if the height of water measured 40 mm in a cylindrical volume, you could state there was 40 mm of rain. If the cone shaped section blocked off part of the cylindrical volume then 40 mm of measured water level would have been created by less than 40 mm of rain. That is the level we are after.**

then

h_cyl = (h_tcone*(r_max^2 + r_max*r_min+r_min^2))/(3*r_cyl^2)

This assume's V truncatedcone=Vcyl - 07 Nov '10 20:26 / 2 edits

im missing something here...this is what im visualizing the problem to be.*Originally posted by sonhouse***Not the total volume of the cone, just the part where the two intersect the 100 mm section of both. The water is unable to fill the whole cylindrical section.**

you are using the truncated cone to as the resivoir to measure the accumulation of rain. You know the crucial dimensions of the of it. You take a meter stick and measure the maximum depth of water inside the truncated conic. This gives you some height. Since the cross-sectional area of the cone changes with height the volume per unit area is not constant...so you need to convert the volume of the truncated conic into a volume of an object with a constant cross-sectional area such as the right circular cylinder.

since the "cone" collected the water from the area of the entire pool, to obtain average rainfall per unit area the "cylinder" has to have a raidius equal to the radius of the pool.

So after "theoretically" taking the water from the truncated cone and pouring it into our "theoretical" cylinder the true accumulation can be measured as the height.

is this what your doing?

If it is, this is the solution

h_cyl= [(h_t-cone)^3 +9*((h_t-cone)^2) + 27*(h_t-cone)]/36.75 - 07 Nov '10 22:32

Sounds like we are getting closer. Just remember, there is a sheet of plastic a half meter down, that defines the lower limit of both the cylinder and the truncated cone so the the water can't fill to the edge of the pool because the cone shaped edge of the plastic sheet prevents water from going to the edge of the pool, so the 100 mm of measured height would be more than the actual rainfall. If the plastic sheet was a real cone, it would go into the ground but it doesn't because of the water left under the plastic sheet makes for a decent level surface but there is an air pocket under the cone shape of the plastic where the angle is 45 degrees from the surface of the water, a half meter below top of pool so the inner radius of the cone section is 6 meters where it hits the pool and the radius of the pool proper is 7 meters. So a triangular section of the cone would be 0.5 meter X 0.5 meter with hypotenuse of 0.707 meter.*Originally posted by joe shmo***im missing something here...this is what im visualizing the problem to be.**

you are using the truncated cone to as the resivoir to measure the accumulation of rain. You know the crucial dimensions of the of it. You take a meter stick and measure the maximum depth of water inside the truncated conic. This gives you some height. Since the cross-sectiona ...[text shortened]... s, this is the solution

h_cyl= [(h_t-cone)^3 +9*((h_t-cone)^2) + 27*(h_t-cone)]/36.75 - 07 Nov '10 23:25

we're not close, the formula I gave should be correct for the specific situation. If you measured 100mm depth in the cone , it corresponds to an actual accumulation of 76 mm over the area of the pool.*Originally posted by sonhouse***Sounds like we are getting closer. Just remember, there is a sheet of plastic a half meter down, that defines the lower limit of both the cylinder and the truncated cone so the the water can't fill to the edge of the pool because the cone shaped edge of the plastic sheet prevents water from going to the edge of the pool, so the 100 mm of measured height wou ...[text shortened]... triangular section of the cone would be 0.5 meter X 0.5 meter with hypotenuse of 0.707 meter.** - 08 Nov '10 00:04 / 10 edits

I took the 40mm depth of the 6 meter diameter and did the PI R^2*H which gave me a volume of 1130973355 cubic mm of water, plus half the volume of a square section 40mmX40mmX6000mmXPI which came out to 30159289.47 cubic mm, added to the previous # to give 1161132644 cubic mm of water total. The ratio of those two #'s is 1.026666 so 40X 1.026 gives 41.06 mm of rain as the true value.*Originally posted by joe shmo***we're not close, the formula I gave should be correct for the specific situation. If you measured 100mm depth in the cone , it corresponds to an actual accumulation of 76 mm over the area of the pool.**

I think you used the 100 mm of ice I mentioned, sorry for the confusion. I should not have rambled on about the ice. The original question was what was the actual rain level. It doesn't matter what the actual diameter of the pool, the cylindrical section that is, a pool 6000 mm in diameter with 40 mm of rain will have a depth of 40 mm if the pool is only 40 mm in diameter assuming the rain is falling uniformly.

So the actual rainfall would be pretty close to the 40 mm measured, but would be 0.76 of that value if it were spread out over a 7000 mm diameter pool or 30.4 mm.

I followed your work and saw that 0.76 figure would be a constant for the conical section as told. So it would be 76 mm deep if the original measurement was 100 mm but 0.76 of the rain measurement of the problem, 40 mm, turning that to 30.4 mm if the cone section were taken out and the water now filled a 7000 mm diameter cylindrical section.

But the level would turn out to be 41.06 mm of actual rain if the plastic was not there or if the sheet conformed exactly to the water line and the side of the pool.

Does my reasoning sound ok to you? - 08 Nov '10 01:20

unfortunately they dont check out...you said in your previous post about the 100 mm of height being measured( i assumed you meant it was the depth of the water in the conic.*Originally posted by sonhouse***I took the 40mm depth of the 6 meter diameter and did the PI R^2*H which gave me a volume of 1130973355 cubic mm of water, plus half the volume of a square section 40mmX40mmX6000mmXPI which came out to 30159289.47 cubic mm, added to the previous # to give 1161132644 cubic mm of water total. The ratio of those two #'s is 1.026666 so 40X 1.026 gives 41.06 mm ...[text shortened]... rmed exactly to the water line and the side of the pool.**

Does my reasoning sound ok to you?

anyhow you'll have to explain this to me

40mmX40mmX6000mmXPI

this is (2*pi*R)*W*H*1/2

but you can't do that. here is why

first of all you used the diameter when based on you calcs above you should have used radius (R=3000mm not 6000mm)

secondly and more importantly, you cant do what you did...As it appears to me you essentialy unwraped the inner circumference and called it "Length"

but you'll notice that there is an outer radius ( coincidently one you do not know, that is based off the height) and the inner circumference does not equal the outer circumference.

If you know the outer circumference than than by the therom of Pappus, you would need to use the average radius.

(2*pi*R_avg)*W*H*1/2

there are some noticable problems, there may be more but you should adress these first. - 08 Nov '10 01:36 / 8 edits

I think there are a couple of ways to do this, one like I said where you calculate the level if it were compared to exactly a section of 6000 mm but it could also be compared to a section of 6000 mm plus 80 mm or 6080 mm. Not sure which solution would be the correct one. I think I am ok in my reasoning though, if the triangular section were forced up to vertical from its original 45 degrees to make the cylindrical section exactly 6000 mm, the water level would rise by 2%. Now I am not so sure. If I did it the other way and compared it to a 6080 mm diameter the water level would go down by 2%. I am visualizing the rain starting with a diameter of exactly 6000 mm but as the rain level rises, it is surely a lower level than if it were a cylindrical section.*Originally posted by sonhouse***I took the 40mm depth of the 6 meter diameter and did the PI R^2*H which gave me a volume of 1130973355 cubic mm of water, plus half the volume of a square section 40mmX40mmX6000mmXPI which came out to 30159289.47 cubic mm, added to the previous # to give 1161132644 cubic mm of water total. The ratio of those two #'s is 1.026666 so 40X 1.026 gives 41.06 mm rmed exactly to the water line and the side of the pool.**

Does my reasoning sound ok to you?

It starts out at 6000 mm in diameter and ends up at 6080 mm in diameter. It seems it should end up how I calculated it the first time.

Why can't I just strip out the triangular section and Pionize it, so to speak, and turn it into a long strip? Oh, I think I see. If I took a 40 mm X 40 mm section and wrapped it around a 6000 mm diameter, the circumference of the resulting section would be different from the circumference of the 6000 mm diameter. But by what percentage?

Since we are comparing a difference of 6000 mm inner diameter at the start of the rain and 6080 mm diameter at the end of the rain, would there be that much difference just using the simplifying assumption I used?

For instance, if you were using a real ruler with one mm spacing, would we be able to see much difference if you compared a real 6000 mm exact cylindrical section compared to my 6000 start and 6080 finish using the assumptions I gave it? In that case I would expect the 6000 mm diameter cylindrical section to be close to my figure of 41 mm.

It seems to me the percentage of error would be greater the smaller the diameter of the cylinder vs the added volume I calculated. For instance if the pool was one kilometer in diameter and the same angle of 45 degrees used, the difference in my method vs the real solution would start to get vanishingly small so 6000 mm vs 6080 mm is not that much different.

If it were 600 mm vs 680 mm, the error would be a lot higher. 6000 to 6080 is only about 1.3% different. 600 vs 680 is about 12% difference. 1,000,000 mm vs 1,000,080 mm is within 0.9999 of each other.

The gist of that is, if the pool is large enough, the bit of difference due to that 45 degree section would pretty much disappear and you could use the measurement as stated by the ruler, but there would be a measurable difference at 6000 mm diameter vs 6080 mm. But by how much using exact methods? - 08 Nov '10 02:42

sonhouse let me spell this out for you*Originally posted by sonhouse***I think there are a couple of ways to do this, one like I said where you calculate the level if it were compared to exactly a section of 6000 mm but it could also be compared to a section of 6000 mm plus 80 mm or 6080 mm. Not sure which solution would be the correct one. I think I am ok in my reasoning though, if the triangular section were forced up to ver ...[text shortened]... be a measurable difference at 6000 mm diameter vs 6080 mm. But by how much using exact methods?**

if you are measureing a depth of 40mm in the conic section

you are summing the volume of the parts of the cone which is fine as long as your volumes are correct, but yours are not!!!

first of all the cylindrical part has a volume of:

V1=pi*(Rmin)^2*h_cone=pi*(3000 mm)^2 * (40 mm) = 1.13097*10^9 mm^3

the volume of the remaining part of the cone is:

broken down

the length is (2*pi***R_avg**) I cant stress enough that**R MUST BE THE AVERAGE RADIUS**

Ill draw a diagram of the crossection (not to scale)

__............__

\..|............./

..\|_______/

R_avg= (3000mm + 3040mm)/2 = 3020 mm

V2= 2*pi*R_avg *W*H*1/2 = pi*3020mm*40mm*40mm=1.51802*10^7 mm^3

V_total=V1+V2 = 1.14615*10^9 mm^3

To get the average rainfall per unit area you need the AREA OVER WHICH THE WATER WAS COLLECTED**WHICH IS THE AREA OF THE ENTIRE POOL**

A_pool=pi*R_pool^2 = pi*(3500mm)^2 =3.84845*10^7 mm^2

V_total/A_pool = (1.14615*10^9 mm^3)/(3.84845*10^7 mm^2) = 29.7822 mm

h_cyl = 29.7822 mm

My Formula

h_cyc=(hcone^3 +9*hcone^2 + 27*hcone)*1000/36.75 = [(.040)^3 +9*(.040)^2+27*(.040)]*1000/36.75 = 29.7813 mm

almost a perfect match (however in the first method (your method) there is more rounding error. - 08 Nov '10 03:00 / 6 edits

Sure, but I am talking about a starting point way up the conic section where it is mostly flat, for instance if you started at the very top of the pool, 7000 mm in diameter and went down 40 mm and used a conic section there, 45 degree angle, the level down there would be 6920 mm. So there would be somewhat less error there than considering a 6000 mm flat circular section with a 45 degree slope starting at the 6000 mm diameter and ending at 6080. It would be like a jet engine that has variable orfice, first it starts out as a full cylindrical section but the 40 mm end can spread out at a 45 degree angle. So the difference would be vanishingly small ( the difference between actual rain level and measured rain level) in a very large pool.*Originally posted by joe shmo***sonhouse let me spell this out for you**

if you are measureing a depth of 40mm in the conic section

you are summing the volume of the parts of the cone which is fine as long as your volumes are correct, but yours are not!!!

first of all the cylindrical part has a volume of:

V1=pi*(Rmin)^2*h_cone=pi*(3000 mm)^2 * (40 mm) = 1.13097*10^9 mm^3

the lmost a perfect match (however in the first method (your method) there is more rounding error.

If the conical section was a lower angle, say 10 degrees, the difference in rain level vs measured level would be a lot more pronounced. Doesn't that sound reasonable?

You know, I get the exact same answer out to 5 digits, if I just calc a 3000 mm radius cylindrical pool 40 mm deep, 1.130973355 E9 mm cubed. Where does the difference start? Oh, I misread that, you DID calc a cylinder. Sorry.

Isn't it safe to assume the volume difference between different radii as PI R^2XH(max R) minus Pi R^2 X H (min R)? So considering a cylindrical section 6080 mm, 3040 mm^2XPiX40mm=116133707 mm^3. 3000 mm^2 X PI X 40=1130973355, subtracting gives you 30,360,352 mm^3 as the difference, and with a 45 degree angle, there would be two volumes of half of that 30 mil figure so the actual volume of that conic section would be 15,180,176 mm cubed, right? I won't go any further than that for now, just want to be sure I am right about that one.