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From: Just Me on 15 May 2010 18:31
Math Whiz needed!

We have a garden pond excavated into a slope and dammed at the
downhill end by a large oak log, 2 feet in diameter and 11 feet long.
On top of the log, I bolted with lag screws a 2 x 8 plank, upon which
is mounted a double stack of concrete blocks. The dam is the Eastern
perimeter, the North and South are composed of rock stacks taken out
of the slope during the excavation which was made easy (if not merry)
by use of a very sturdy old Murray garden tiller, built with tines of
steel heavy gauge enough to withstand anything this rocky, rooted
ground can offer it.

The lengths of the North and South perimeters is irregular and so I
figure to make this easy by giving the average of the two as 9.5
feet. So we have a rectangle of 9.5 x 11 feet.

The water at its present level is given here also as an average of the
soundings taken with a yard-stick, yielding an average depth along the
dam of exactly 2 feet. The average depth on the opposite East side
is 15 inches.

If I were a mathematician, I could now discover the angle of the slope
under the pond by a calculation of those two depth average figures.
But I am not a mathematician which is why I am here, specifically to
discover approximately what would be the volume of water in my pond in
figures of both cubic feet and gallons.

Who here is might enough mathematically to tell me how much rain-water
I have captured from the storms of this past week?
--
JM
From: jmorriss on 15 May 2010 22:42
On May 15, 6:31 pm, Just Me <jpd...(a)gmail.com> wrote:
> Math Whiz needed!
>
> We have a garden pond excavated into a slope and dammed at the
> downhill end by a large oak log, 2 feet in diameter and 11 feet long.
> On top of the log, I bolted with lag screws a 2 x 8 plank, upon which
> is mounted a double stack of concrete blocks. The dam is the Eastern
> perimeter, the North and South are composed of rock stacks taken out
> of the slope during the excavation which was made easy (if not merry)
> by use of a very sturdy old Murray garden tiller, built with tines of
> steel heavy gauge enough to withstand anything this rocky, rooted
> ground can offer it.
>
> The lengths of the North and South perimeters is irregular and so I
> figure to make this easy by giving the average of the two as 9.5
> feet.  So we have a rectangle of 9.5 x 11 feet.
>
> The water at its present level is given here also as an average of the
> soundings taken with a yard-stick, yielding an average depth along the
> dam of  exactly 2 feet.  The average depth on the opposite East side
> is 15 inches.
>
> If I were a mathematician, I could now discover the angle of the slope
> under the pond by a calculation of those two depth average figures.
> But I am not a mathematician which is why I am here, specifically to
> discover approximately what would be the volume of water in my pond in
> figures of both cubic feet and gallons.
>
> Who here is might enough mathematically to tell me how much rain-water
> I have captured from the storms of this past week?
> --
> JM

Length times Width times Average Depth... Make sure your units are
consistent...
From: Cwatters on 16 May 2010 04:56

"Just Me" <jpdm45(a)gmail.com> wrote in message
news:8c2b7c29-fe1e-4a22-a7f8-19b1ebd58b45(a)e28g2000vbd.googlegroups.com...
> Math Whiz needed!
>
> We have a garden pond excavated into a slope and dammed at the
> downhill end by a large oak log, 2 feet in diameter and 11 feet long.
> On top of the log, I bolted with lag screws a 2 x 8 plank, upon which
> is mounted a double stack of concrete blocks. The dam is the Eastern
> perimeter, the North and South are composed of rock stacks taken out
> of the slope during the excavation which was made easy (if not merry)
> by use of a very sturdy old Murray garden tiller, built with tines of
> steel heavy gauge enough to withstand anything this rocky, rooted
> ground can offer it.
>
> The lengths of the North and South perimeters is irregular and so I
> figure to make this easy by giving the average of the two as 9.5
> feet. So we have a rectangle of 9.5 x 11 feet.
>
> The water at its present level is given here also as an average of the
> soundings taken with a yard-stick, yielding an average depth along the
> dam of exactly 2 feet. The average depth on the opposite East side
> is 15 inches.
>
> If I were a mathematician, I could now discover the angle of the slope
> under the pond by a calculation of those two depth average figures.
> But I am not a mathematician which is why I am here, specifically to
> discover approximately what would be the volume of water in my pond in
> figures of both cubic feet and gallons.
>
> Who here is might enough mathematically to tell me how much rain-water
> I have captured from the storms of this past week?
> --
> JM

Working in inches..
Average depth is (24+15)/2 = 19.5 inches
Area is 9.5 x 12 x 11 x 12 = 15048 square inches
Volume = 19.5 x 15048 = 293436 cubic inches

Convet to cubic feet..
293436 / (12 x 12 x 12) = 169.8 cubic feet

Google says that's 1270 US Gallons.


From: Just Me on 18 May 2010 02:29
On May 15, 9:42 pm, "jmorr...(a)idirect.com" <jmorr...(a)idirect.com>
wrote:
> On May 15, 6:31 pm, Just Me <jpd...(a)gmail.com> wrote:
>
>
>
>
>
> > Math Whiz needed!
>
> > We have a garden pond excavated into a slope and dammed at the
> > downhill end by a large oak log, 2 feet in diameter and 11 feet long.
> > On top of the log, I bolted with lag screws a 2 x 8 plank, upon which
> > is mounted a double stack of concrete blocks. The dam is the Eastern
> > perimeter, the North and South are composed of rock stacks taken out
> > of the slope during the excavation which was made easy (if not merry)
> > by use of a very sturdy old Murray garden tiller, built with tines of
> > steel heavy gauge enough to withstand anything this rocky, rooted
> > ground can offer it.
>
> > The lengths of the North and South perimeters is irregular and so I
> > figure to make this easy by giving the average of the two as 9.5
> > feet.  So we have a rectangle of 9.5 x 11 feet.
>
> > The water at its present level is given here also as an average of the
> > soundings taken with a yard-stick, yielding an average depth along the
> > dam of  exactly 2 feet.  The average depth on the opposite East side
> > is 15 inches.
>
> > If I were a mathematician, I could now discover the angle of the slope
> > under the pond by a calculation of those two depth average figures.
> > But I am not a mathematician which is why I am here, specifically to
> > discover approximately what would be the volume of water in my pond in
> > figures of both cubic feet and gallons.
>
> > Who here is might enough mathematically to tell me how much rain-water
> > I have captured from the storms of this past week?
> > --
> > JM
>
> Length times Width times Average Depth...  Make sure your units are
> consistent...

Thanks!
--JM
From: Just Me on 18 May 2010 02:30
On May 16, 3:56 am, "Cwatters"
<colin.wattersNOS...(a)TurnersOakNOSPAM.plus.com> wrote:
> "Just Me" <jpd...(a)gmail.com> wrote in message
>
> news:8c2b7c29-fe1e-4a22-a7f8-19b1ebd58b45(a)e28g2000vbd.googlegroups.com...
>
>
>
>
>
> > Math Whiz needed!
>
> > We have a garden pond excavated into a slope and dammed at the
> > downhill end by a large oak log, 2 feet in diameter and 11 feet long.
> > On top of the log, I bolted with lag screws a 2 x 8 plank, upon which
> > is mounted a double stack of concrete blocks. The dam is the Eastern
> > perimeter, the North and South are composed of rock stacks taken out
> > of the slope during the excavation which was made easy (if not merry)
> > by use of a very sturdy old Murray garden tiller, built with tines of
> > steel heavy gauge enough to withstand anything this rocky, rooted
> > ground can offer it.
>
> > The lengths of the North and South perimeters is irregular and so I
> > figure to make this easy by giving the average of the two as 9.5
> > feet.  So we have a rectangle of 9.5 x 11 feet.
>
> > The water at its present level is given here also as an average of the
> > soundings taken with a yard-stick, yielding an average depth along the
> > dam of  exactly 2 feet.  The average depth on the opposite East side
> > is 15 inches.
>
> > If I were a mathematician, I could now discover the angle of the slope
> > under the pond by a calculation of those two depth average figures.
> > But I am not a mathematician which is why I am here, specifically to
> > discover approximately what would be the volume of water in my pond in
> > figures of both cubic feet and gallons.
>
> > Who here is might enough mathematically to tell me how much rain-water
> > I have captured from the storms of this past week?
> > --
> > JM
>
> Working in inches..
> Average depth is (24+15)/2 = 19.5 inches
> Area is  9.5 x 12 x 11 x 12 = 15048 square inches
> Volume = 19.5 x 15048 = 293436 cubic inches
>
> Convet to cubic feet..
> 293436 / (12 x 12 x 12) = 169.8 cubic feet
>
> Google says that's 1270 US Gallons.

Thanks!
--
JM
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