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bcassel
02-25-2015, 08:10 PM
I have acquired an Alemite oil dispenser and am in the process of rehabing it. But I'm stumped on the volume that this dispenser will hold. It is a funnel shape. I'll include my calculations and, if you would, tell me if I'm on target. A photo is also attached.

The tank is 16" tall (H)
The top is an oval, 12" (A) by 7.5" (B)
The bottom is a circle, 6" in diameter (C)
A gallon is 231 cubic inches

So, I figure to determine the volume, I calculate the area of the top oval and the bottom circle. Average these to determine the average area.

I multiply the average area by the height to determine cubic inches of volume.

I divide the volume cubic inches by 231 to determine the gallons this will hold.

So:
area of top oval = pi * A * B 3.14 * 12 * 7.5 = 282.6 sq inches
area of bottom circle = pi * C/2 squared = 28.26 square inches
average of the two = 310.86 cubic inches/2 = 155.43
to compute total volume, average of area (155.43) time H (16") = 2486.88 cubic inches
convert cubic inches to gallons = 2486.88 / 231 =10.7

Am I missing something? It doesn't look that big!

http://bbs.homeshopmachinist.net/album.php?albumid=71&attachmentid=548

http://bbs.homeshopmachinist.net/album.php?albumid=71&attachmentid=548

bcassel
02-25-2015, 08:11 PM
Here is the link to the photo:

http://bbs.homeshopmachinist.net/album.php?albumid=71&attachmentid=548

oldtiffie
02-25-2015, 08:51 PM
Why not leave it empty and fill it from/by a graduated jug/container?

You might like to use the same process to find the oil discharge per stroke of the pump as well while you are at it.

Doozer
02-25-2015, 09:21 PM
Fill it with water and weigh it??

-Doozer

02-25-2015, 09:36 PM
I got a top area of 70.7 in.sq. I drew it up in TurboCAD and got a volume of about 3 1/3 gallons.

oldtiffie
02-25-2015, 09:52 PM
Fill it with water and weigh it??

-Doozer
Its easier in metric as 1 litre = 1 Kilogram

Stepside
02-25-2015, 10:06 PM
I drew it in Rhinoceros Assuming the top is a true ellipse and the sides are a straight transition from round to ellipse. I got 720.152 inches cubed which divided by 231 equals 3.117 gallons. I suspect it was designed as a 3 gallon container.

cameron
02-26-2015, 12:21 AM
Its easier in metric as 1 litre = 1 Kilogram

And 1 Imperial gallon of water weighs 10 pounds, how easy does it have to be? Or I cubic foot weighs 62.4 pounds. If you used the Imperial system you would know those numbers, and you can punch them into a cheap calculator, if you can't manage with a pencil and something to write on.

bcassel
02-26-2015, 01:25 AM
I found my error! I did not halve the length and width of the top oval before I put them into the equation. Duh! Correct answer is 3.4 gallons, thanx for those who helped

Paul Alciatore
02-26-2015, 02:58 AM
The shape you have is similar to a cone, but not identical so determining the exact volume will be difficult.

If it were a cone, even then your procedure would not be correct. The diameter would change in a linear manner from the top circle to the bottom circle, but the area would change by the square of that linear dimension. So taking an average will not work. You need more complicated math. I would do it, but since you do not have a circle at the top, it would still be wrong. Even with the major and minor diameters of that oval or ellipse, you still do not know it's exact area because you do not know which it is: oval, ellipse, or something else. And then, how does it transition from that unknown shape to the circle at the bottom? That could vary and is also unknown.

Due to all of this, your calculated number is too high. But just how much, I do not know.

Your best bet would be to just fill it with water and then pour that water into a calibrated container.

I have acquired an Alemite oil dispenser and am in the process of rehabing it. But I'm stumped on the volume that this dispenser will hold. It is a funnel shape. I'll include my calculations and, if you would, tell me if I'm on target. A photo is also attached.

The tank is 16" tall (H)
The top is an oval, 12" (A) by 7.5" (B)
The bottom is a circle, 6" in diameter (C)
A gallon is 231 cubic inches

So, I figure to determine the volume, I calculate the area of the top oval and the bottom circle. Average these to determine the average area.

I multiply the average area by the height to determine cubic inches of volume.

I divide the volume cubic inches by 231 to determine the gallons this will hold.

So:
area of top oval = pi * A * B 3.14 * 12 * 7.5 = 282.6 sq inches
area of bottom circle = pi * C/2 squared = 28.26 square inches
average of the two = 310.86 cubic inches/2 = 155.43
to compute total volume, average of area (155.43) time H (16") = 2486.88 cubic inches
convert cubic inches to gallons = 2486.88 / 231 =10.7

Am I missing something? It doesn't look that big!

http://bbs.homeshopmachinist.net/album.php?albumid=71&attachmentid=548

http://bbs.homeshopmachinist.net/album.php?albumid=71&attachmentid=548

Paul Alciatore
02-26-2015, 03:09 AM
If the top really is a true oval that consists of two half round ends with 7.5" diameters and a central rectangle that measures 7.5" X (12"-7.5"), then it's area is given by:

A = [Pi x (7.5/2)^2] + [7.5 x 4.5]

A = [3.141 x [7.5 x 4.5]

A = 56.18 sq in

You use the radius squared, not the diameter squared.

But I suspect your corrected result of 3.4 gallons is still too large because you can not just average the two areas.

peekaboobus
02-26-2015, 04:52 AM
And 1 Imperial gallon of water weighs 10 pounds, how easy does it have to be? Or I cubic foot weighs 62.4 pounds. If you used the Imperial system you would know those numbers, and you can punch them into a cheap calculator, if you can't manage with a pencil and something to write on.

I can't manage arithmetic either. And my calculator is not cheap.

rock_breaker
02-26-2015, 03:16 PM
I might be missing something here as well, but in your text you multiply the top dimensions by 12, are they not already in inches? My point is mute now as others have come up with what I believe to be correct answers but I think I would have averaged the top dimensions. In conclusion the points made about measuring are well taken.

Have a good day

Ray

CCWKen
02-26-2015, 04:17 PM
I have the same can but with a manual pump. It's very old. I haven't had a chance to clean it up and take some dents out of it. Yours looks in great shape. Since I do antique cars, I thought it would be a nice accessory for the shop.

Barrington
02-27-2015, 05:50 AM
If anyone's interested in how to calculate the volume of such a shape:-

Assuming the top is elliptical, the circular bottom is also just a special case of an ellipse, so,

The top has a semi-minor axis 'a' of 3.75", and a semi-major axis 'b' of 6"
The bottom has a semi-minor axis 'a' of 3" and a semi-major axis 'b' also of 3"

'a' then goes from 3" to 3.75" over a height 'h' of 16", i.e. a = 3 + 0.75h/16
'b' then goes from 3" to 6" over a height 'h' of 16", i.e. b = 3 + 3h/16

The cross sectional area at height 'h' = pi * a * b

To save time (i.e. being lazy), use 'Microsoft Mathematics' (a free download) to do the integration:-

http://i564.photobucket.com/albums/ss82/MrBarrington/volume_zpsn95ot6sh.png

772.83 cubic inches = 2.79 proper gallons :p or 3.35 little US gallons.

Cheers

.