- posted on September 17, 2005, 4:26 pm

My shop is heated with an old oil burner. The tank is a 38" by 60" cylinder which I believe is 300 gallons. Because the tanks sits on a stand on its side I need to know how to convert the inches remaining in the tank to gallons. The input pipe is on top and currently shows 15" on the stick. Obviously 19" would equate to 150 gallons but how do I graduate the measure stick to measure more than half as well as less than half? Or perhaps the better question is how do I create a chart that equates inches to gallons?

Patrick Fischer Olalla, WA

- posted on September 17, 2005, 5:49 pm

First step is to go tot he Lee Valley web site and order the Pocket Reference by T J Glover. Then go to page 422 and find the ratio then the depth factor for your tank. For a 38" tank the ratio is H/D or .39, rounded to.40 thus the depth factor is 0.3735300. depth factor x total gallons 112 gallons. This is assuming you are giving inside dimensions, not outside.

The Pocket Reference is a very handy book I use it frequently for some obscure formula that comes up. It also give floor loading for joists, airport codes, flow through pipes, steam tables, and lots of other electrical, mechanical, chemical and structural information.

--

Ed

http://pages.cthome.net/edhome/

Ed

http://pages.cthome.net/edhome/

- posted on September 18, 2005, 12:05 am

rounded

Looks like you just gave a linear equation for a tank on its end. His tank is on its side and the equation is non-linear..

Bob

- posted on September 18, 2005, 4:03 am

No shit, that is why I chose this formula. Page 422 is titled Horizontal Cylinder Fillage

It states: "The following equation can be used to calculate the number of gallons remaining in a horizontal tank if the height of the liquid remaining in the tank and the diameter of the tank are known."

You did look at the page didn't you?

- posted on September 19, 2005, 1:52 am

wrote in message

remaining

Edwin, don't get huffy. No I didn't look at the page. I know enough to know you cannot condense volume to a linear multiplication for this example, I don't care what the book says. One of the other posts showed a non-linear equation which it appeared correct, but I did not work through it to see if I thought it was correct.

Bob

remaining

Edwin, don't get huffy. No I didn't look at the page. I know enough to know you cannot condense volume to a linear multiplication for this example, I don't care what the book says. One of the other posts showed a non-linear equation which it appeared correct, but I did not work through it to see if I thought it was correct.

Bob

- posted on September 19, 2005, 2:53 am

Then look at the book. They give a depth factor depending on the ratio. Rather than make comments about something you have never seen, check it out and you will find it to be correct. To understand this you really have to LOOK AT THE BOOK. Page 422. I got mine from Lee Valley, but it is sold elsewhere. Sorry, but your opinion is superceded by facts.

- posted on September 17, 2005, 6:05 pm

On Sat, 17 Sep 2005 09:26:55 -0700, Patrick Fischer wrote:

d, r, L in inches. d is measured depth. r is radius of tank. L is length of tank. V in gallons (231 cu in per gal) f(d) = 1/2 (((r^2)*arcsin(1/r)) - (r-d)*sqrt(r^2 - (r-d)^2) + r^2 * arcsin((r-d)/r^2)

*d>r; V= ((f(d) + 1/2****pi***r^2)*L)/231
d<r; V= (f(d) * L) /231
d=r; V= (1/2 *** pi *** r^2 * L)/231

I took calculus thirty years ago. Someone a little more current should check this... :)

d, r, L in inches. d is measured depth. r is radius of tank. L is length of tank. V in gallons (231 cu in per gal) f(d) = 1/2 (((r^2)*arcsin(1/r)) - (r-d)*sqrt(r^2 - (r-d)^2) + r^2 * arcsin((r-d)/r^2)

I took calculus thirty years ago. Someone a little more current should check this... :)

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- posted on September 17, 2005, 8:01 pm

Every oil company I have ever used has sticks with that info printed on it. You stick the stick in and read off where the wil is on the stick. then wipe it off with a rag.

Justmake sure you get the right stick. Just ask for one. They like giving them out because it encourages oil sales.

- posted on September 17, 2005, 11:58 pm

On Sat, 17 Sep 2005 16:01:03 -0400, "Lee Michaels"

I missed the OP, but solved that problem as a kid, using high school math. The result is sinusoidal, and not simple for anyone not familiar with the math. Basically, you draw a circle with a chord [initially below the half-way diameter for convenience] and draw the sector. The area of the lower segment [the liquid you want to find] made by the chord is the area of the sector less that of the triangle formed by the two arms of the sector [radii] and the chord. The first is related to the sector angle in radians, and the second is the familiar triangle area. However, tha uses th chord length, so to tie the two together you have to use Pythagoras' theorem. You have to express the angle as an inverse trig function [why the "sinusoidal" reference] in terms of the tank radius R, and the liquid height, h.

If you drew the half sector, you'd have an angle,A, a radius R off to the right to meet the chord at the circle, a liquid height h, and a remaining height of R-h, the third sied of a triangle of side R and c/2, half the chord length.

Volume will be the resulting area times the lenght of the container.

...you really don't want me to go on with all of this, but you did ask. Also, it's a bit difficult to try to write it all in plain text, so the best bet is to look it up in a web search, or some possible boilermaker tables you might find in a used book store.

I missed the OP, but solved that problem as a kid, using high school math. The result is sinusoidal, and not simple for anyone not familiar with the math. Basically, you draw a circle with a chord [initially below the half-way diameter for convenience] and draw the sector. The area of the lower segment [the liquid you want to find] made by the chord is the area of the sector less that of the triangle formed by the two arms of the sector [radii] and the chord. The first is related to the sector angle in radians, and the second is the familiar triangle area. However, tha uses th chord length, so to tie the two together you have to use Pythagoras' theorem. You have to express the angle as an inverse trig function [why the "sinusoidal" reference] in terms of the tank radius R, and the liquid height, h.

If you drew the half sector, you'd have an angle,A, a radius R off to the right to meet the chord at the circle, a liquid height h, and a remaining height of R-h, the third sied of a triangle of side R and c/2, half the chord length.

Volume will be the resulting area times the lenght of the container.

...you really don't want me to go on with all of this, but you did ask. Also, it's a bit difficult to try to write it all in plain text, so the best bet is to look it up in a web search, or some possible boilermaker tables you might find in a used book store.

- posted on September 18, 2005, 1:46 am

Thanks to all who responded. The "solution" turns out to be rather easy. The guy at the fuel store gave me a chart.

Patrick Fischer Olalla, WA

- posted on September 18, 2005, 4:18 am

On Sat, 17 Sep 2005 18:46:44 -0700, "Patrick Fischer"

i.e., somebody else already did the calculus and generated a chart. This could be a really interesting problem.

+--------------------------------------------------------------------------------+ If you're gonna be dumb, you better be tough +--------------------------------------------------------------------------------+

i.e., somebody else already did the calculus and generated a chart. This could be a really interesting problem.

+--------------------------------------------------------------------------------+ If you're gonna be dumb, you better be tough +--------------------------------------------------------------------------------+

- posted on September 18, 2005, 6:12 pm

On Sat, 17 Sep 2005 21:18:27 -0700, Mark & Juanita wrote:

Yeah. I've realized that I had a factor-of-two error in the solution I posted earlier. Geometry, trig, and calc are useful shop skills. Fetching the chart from a vendor is another useful skill...

Yeah. I've realized that I had a factor-of-two error in the solution I posted earlier. Geometry, trig, and calc are useful shop skills. Fetching the chart from a vendor is another useful skill...

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- posted on September 18, 2005, 11:02 pm

On Sat, 17 Sep 2005 09:26:55 -0700, "Patrick Fischer"

There are approximately 231 cubic inches per gallon. Or, 7.481 gallons per cubic foot. To calculate the volume of a cylinder multiply the base area by the height.

There are approximately 231 cubic inches per gallon. Or, 7.481 gallons per cubic foot. To calculate the volume of a cylinder multiply the base area by the height.

- posted on September 19, 2005, 12:52 am

On Sun, 18 Sep 2005 23:02:50 +0000, Phisherman wrote:

But OP wants to know how much oil is left. Yes, pi*r^2*h gives the total volume. It's trickier when the cylinder is on its side and you want to know the volume of oil in the tank, not the tank's capacity. Calculus (or the vendor's chart) comes to the rescue.

But OP wants to know how much oil is left. Yes, pi*r^2*h gives the total volume. It's trickier when the cylinder is on its side and you want to know the volume of oil in the tank, not the tank's capacity. Calculus (or the vendor's chart) comes to the rescue.

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- posted on September 19, 2005, 1:07 am

Or CAD.

. Calculus (or

. Calculus (or

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