Amount - Gallons in Tank

[Guest]

A railroad tank car is stationary on a track and the amount of gasoline in the tank is measured. A stick placed through an opening in the top of the tank determines the gasoline is 15 inches high.

Dimensions of tank:

Diameter - 60 inches.
Length- 25 feet.

Calculate amount (gallons) in tank.


My partial solution:

Volume of cylinder - pi x radius squared x height (or length).

3.1416 x 30 x 30 x 300 (25 feet converted to inches) = 848232 cu. in.

848232 is total volume of tank - The contents are 15 inches in height.

How do I calculate the amount in tank ?

[Guest]

Assuming the tank is on it's side (so the circular ends lie in a vertical plane) then you need to calculate the area of a segment of a circle and use that to calculate the volume.

The area of a segment of a circle can be calculated by finding the area of a sector and subtracting the area of an isosceles triangle. There are lots of tutorials and videos online which show you how to do this if you know the angle of the sector. You can work out the angle of the sector by using trigonometry on the isosceles triangle (split it along its symmetry line to get two right angled triangles).

http://www.mathopenref.com/cylindervolpartial.html
http://www.mathopenref.com/segmentareaht.html

The second link gives you a formula for the area of a segment. Using trigonometry on half of the isosceles triangle they can work out that that one of the angles, lets call it [tex]\theta[/tex], satisfies [tex]\cos \theta = (r-h)/r[/tex] (note that [tex]r-h[/tex] is the height of the triangle, [tex]r[/tex] is the hypotenuse). The angle of the sector is [tex]2\theta[/tex] or by substituting the previous formula we get [tex]2\cos^{-1}((r-h)/r)[/tex]. If we take the result in radians then the proportion of the circle this represents is [tex]2\cos^{-1}((r-h)/r) / 2\pi = \cos^{-1}((r-h)/r) / \pi[/tex], since the area of the entire circle is [tex]\pi r^2[/tex] the area of the sector is [tex]\pi r^2 \cos^{-1}((r-h)/r) / \pi = r^2\cos^{-1}((r-h)/r)[/tex]. Recall that we split the isosceles triangle into two right angled triangles, and that triangle had height [tex]r-h[/tex] and hypotenuse [tex]r[/tex]. By Pythagoras it must have a base of [tex]\sqrt{r^2-(r-h)^2} = \sqrt{r^2-r^2+2hr-h^2} = \sqrt{2hr-h^2}[/tex]. So the area of the right angle triangle is 1/2 times base times height [tex]=1/2\times (r-h)\sqrt{2hr-h^2}[/tex], and the area of the isosceles triangle is [tex](r-h)\sqrt{2hr-h^2}[/tex]. Subtracting the area of the segment from the area of the isosceles triangle gives the area of the segment
[tex]r^2\cos^{-1}((r-h)/r) - (r-h)\sqrt{2hr-h^2}[/tex]
(Multiply this by the length of the tank to get the volume.)

Hope this helped,

R. Baber.

[Guest]

I didn't take a trigonometry course so I don't understand your solution. I'll read the information on the links you provided and more to get an understanding. I didn't know the solution involves trig. Thanks again.

[Guest]

My last line is wrong. It should read:
"Subtracting the area of the isosceles triangle from the area of the sector gives the area of the segment"

R. Baber.

[Guest]

Maybe if we just work it out and keep it simple it may be easier.......

Depth of oil = 15 inches = 1.25 ft
Diameter = 60 inches = 5 ft ...Radius = 2.5 ft
so dist from oil to centre = 2.5 - 1.25 = 1.25 ft

Angle of sector subtended from centre to the edge of the oil surface
= 2 x (ArcCos 1.25/2.5) = 2 x 60 degrees = 120 degrees

Area of sector = (120/360) x Pi x 2.5^2 = 1/3 x Pi x 6.25 = 6.54 sq ft

Area of triangle bit of sector = (dist oil to centre) x (half width of oil)

half width of oil (Pythagoras) = sqrt(2.5^2 - 1.25^2) = sqrt(3.75 x 1.25) = 2.165 ft

So Area of triangle bit of sector = 1.25 x 2.165 = 2.706 sq ft

Area of segment of oil = 6.54 - 2.706 = 3.83 sq ft.

Vol of oil = 3.83 x 25 = 95.75 cu ft.

1 cu ft = 7.48 US liquid gallons

So Vol of oil = 7.48 x 95.75 = 716.21 US liquid gallons gasoline in tank

[Guest]

" Angle of sector subtended from centre to the edge of the oil surface
= 2 x (ArcCos 1.25/2.5) = 2 x 60 degrees = 120 degrees

Area of sector = (120/360) x Pi x 2.5^2 = 1/3 x Pi x 6.25 = 6.54 sq ft

Area of triangle bit of sector = (dist oil to centre) x (half width of oil) "

I don't understand these steps.

[Guest]

" Angle of sector subtended from centre to the edge of the oil surface "

The question tells us that the oil is 15 inches or 1.25 feet deep.
We are not told the tank is cylindrical but we have assumed that from the fact that we are given a diameter and a length. The fact we are given the length implies that the length is placed horizontal rather than a vertical height. So we assume the tank is circular in cross-section and lying horizontal with 15 inches deep of oil lying in the curved bottom. The oil will have a natural flat top surface and so forms a segment of the circular cross-section of the tank. The top surface of the oil forms a chord of the circle and extends to the curved edge of the tank. If we imaging a line joining the ends of this chord to the centre point of the circle ( in fact 2 radii ), this encloses the trianglular part of a sector of the circle. The angle subtended by these 2 radii is the angle of the sector of the circle and we found this to be 120 degrees.......because the vertical distance from the centre to the oil surface is 1.25 and this divided by the radius (hypotenuse) is the cosine of half of the sector angle....so that gives us ....that twice that will be the whole angle so we get...Angle of the sector = 2 x (ArcCos 1.25/2.5) = 2 x 60 degrees = 120 degrees ...where (ArcCos 1.25/2.5) means the inverse of Cos OR the angle that has the Cos(1.25/2.5) OR Cos(0.5) .....


"Area of sector = (120/360) x Pi x 2.5^2 = 1/3 x Pi x 6.25 = 6.54 sq ft"

A sector is a part of a circle enclosed by an arc and 2 radii from the centre. It comprises of a triangle bit made up of 2 radii and a chord (actually the oil surface in this case) and a segment bit bounded by the oil surface and the curvrd arc of the tank.

The area of a complete circle is Pi x r^2. A complete circle is 360 degrees.
The sector we have has an angle ( at the centre) of 120 degrees .... this is 120/360 = 1/3 of a circle...so the sector area will be 1/3 of a circle area.
And so we get..... 1/3 x Pi x 6.25 = 6.54 sq ft


Area of triangle bit of sector = (dist oil to centre) x (half width of oil) "

You can imagine the triangular bit cut vertically into two pieces ... as 2 identical right-angled triangles that can be placed together to form a rectangle. The height will be the distance from the centre of the tank to the oil surface and the width of the rectangle will be half the width of the oil surface. We use Pythagoras to find half the width of the oil by sqrt(2.5^2 - 1.25^2) that is the hypotenuse (radius) squared minus the vertical distance (centre to oil) squared and take the square root of it.
Then we multiply the height by the width to get the area ...this is the area of the triangular bit of the sector.

Then we subtract the triangular area from the sector area to be left with the segment bit area of the oil. Then multiply by the length of the tank and convert to gallons.

[Guest]

Thanks - I am not totally clear on your explanation but I understand more than I did.

[Guest]

Draw a diagram of a cross-section of the tank. A circle with a line to show the level of oil and show the centre of the circle. Draw radius line from centre of circle to each end of the oil level line. This then shows a sector of the circle made up of a triangular bit and a segment which is the area oil the oil.

Let D = 5 = diameter...so.. Radius r is 2.5 feet
Let h = height of oil level from bottom = 15 inches or 1.25 feet
Let w = half the width of the oil level.

Then (r^2 - (r-h)^2) = W^2 OR W = Sqroot(r^2 - (r-h)^2) = 2.165 feet
Then area of the triangle is the height (r-h) times the half width [area of a triangle is (height x base/2].
So that is 1.25 x 2.165 = 2.706 sq feet.

The angle of the sector is 2( arcCos(r-h)/r ) = 2 x arcCos0.5 = 2 x 60 = 120 degrees = 120/360 = 1/3 of a circle.
Area of sector is 1/3 Pi x r^2 = 1/3 x 3.142 x 2.5 x 2.5 = 6.54 sq ft.

Area of segment of oil = 6.54 - 2.706 = 3.83 sq ft.

Vol of oil = 3.83 x 25 = 95.75 cu ft.

1 cu ft = 7.48 US liquid gallons

So vol of oil = 7.48 x 95.75 = 716.21 US liquid gallons gasoline in tank

[Guest]

I'll do that. Thanks.

[Guest]

If you study the diagram drawn and imagine or draw a vertical line from the centre to touch the oil surface line, this splits the triangle into 2 right angled triangles. The vertical distance (r-h) = 1.25 and the hypotenuse is the radius = 2.5. We see that these sides are in the ratio of 1 : 2 so the third side (the half width of the oil line) must be in the ratio of Sqroot of 3 to agree with Pythagoras (1 : 2 : sqrt3).

So half width of oil is 1.732 x 1.25 = 2.165 feet.

The area of the sector triangle bit is then 1.25 x 2.165 = 2.706 sq feet.

Also a 1 : 2 : sqroot3 triangle is a 60 degree, 30 degree right angled triangle so the half sector angle at the centre is 60 degrees giving 2 x 60 = 120 degrees for the sector angle and this is 1 /3 of a circle.

So Area of sector is 1/3 Pi x r^2 = 1/3 x 3.142 x 2.5 x 2.5 = 6.54 sq ft

Area of segment of oil = 6.54 - 2.706 = 3.83 sq ft.

Vol of oil = 3.83 x 25 = 95.75 cu ft.

1 cu ft = 7.48 US liquid gallons

So vol of oil = 7.48 x 95.75 = 716.21 US liquid gallons gasoline in tank.

So we have solved the problem using the basic properties of simple circles and triangles and did not need to do extended algebriac or trigonometry calculations......because the figures given were simple.........

[Guest]

Thanks.

I still don't understand the arcCos.

[Guest]

I still don't understand the arcCos.
The Cos of an angle in a right angled triangle is the Cosine function of the angle given by the side adjacent to the angle divided by the hypotenuse
Cos = A/H .

OR you can look up the Cosine of an angle in your calculator.....

The Cos of 60 degrees ( Cos60 ) is 0.5

If you want to do the reverse and find what angle corresponds to a particular value for Cosine eg if A/H = 0.5 then you would look up the ArcCos of 0.5. Some books call it the Inverse Cos function OR Cos to the minus 1. I cannot type Cos to the minus 1 here except using Tex ???? so I used ArcCos instead

[Guest]

Thanks.

[Guest]

As a matter of curiosity..... Working in radians instead of degrees............

Radius = 2.5 ft ........... Oil depth = 1.25 ft

Centre to oil dist. = (2.5 - 1.25) = 1.25 ft ...... all as before .........

Sector half angle = ArcCos(0.5) = 1.047 radians

Whole Sector Area = Radius^2 x halfAngle = 2.5^2 x 1.047 = 6.544 sq ft

Whole Sector Triangle bit Area = R x Cos(1.047) x R x Sin(1.047) = 2.5 x 0.5 x 2.5 x 0.866 = 2.706 sq ft

The rest is as before..............

[Guest]

Thanks for additional info.

[Guest]

The last post is based on half the sector angle.

If we base calculations on the whole sector angle at the centre then.....and still in radians.....

Area of sector is R^2 x Angle / 2

If we take the triangular bit as having a base = R and a perp. height = R x Sin(Angle) then ....
Area of triangle bit = R x R x Sin(Angle) / 2

This simplifies to ... [ (R^2) / 2 ] x [ (Angle) - Sin(Angle) ] for the Area of the segment.

Sector full centre angle = 2 x ArcCos(0.5) = 2.094 radians

So area of segment = [ 6.25 / 2 ] x [ 2.094 - 0.866 ] = 3.837 sq ft. ..... as before .....

[Guest]

Thanks for additional - not totally clear.

[Guest]

Please reply - Thanks.

[Guest]

Never mind. I don't understand the solution. I thought I did after the drawing but I don't. Sorry I wasted your time. I didn't know the solution was so complicated. Thanks anyway.