Here’s the version solved with calculus, as Ray suggested. I got the same answer of 5.96 inches (.298 of the way up a 20-inch pole) as many others. And, of course, the 3/4 mark is 10 + 5.96 inches up from the bottom.
This solution uses a little trig and then iterates for the angle. No integral calculus is used. See the attached file.
E. Terry Schulze
I almost did it right. r=10, not 20. So the answer is 4.24 inches above and below center.
Dear Tom and Ray:
When you call Rich back to tell him the answer (or when you e-mail him), please don’t call his fuel “gas.” This is a pet peeve of many truck drivers. They don’t use gasoline in their fuel tanks.
Oops - 3/4 mark was wrong (it was late at night when I posted it). 3/4 mark is actually 20 inches minus the 5.96 inches.
Everyone seems to want to argue about math and not actually solve the problem in the most simple and accurate and least inconvenient way possible (esp. for a trucker).<.blockquote>
The least inconvenient way for a trucker to figure this out is to call in to a radio show and let them, or dozens of their listeners, figure it out.
It’s simple Fill Your tanks. You should know what your average fuel milage is. For example if you get 7.0 mpg and you carry 300 gal when full, Run 525 miles stick your tanks mark your stick drive onother 525 miles stick your tanks drive another 525 miles stick your tanks that gives you 1/4 1/2 and 3/4 marks. The easy just drive bi milage fill your tanks every time drive 1575 miles you have used 3/4 of your fuel.
Painful as it may be, Tommy was on the right track using volume. No calculus required. Rich already has a wooden dowel for measuring a half tank.
First, as Tommy suggested, ascertain the volume of the fuel tank. (pi) x (radius squared) x (height).
For sake of this example, radius is 10 inches and assume the height to be 29.4 inches.
Volume = pi x 100 square inches x 29.4 inches = 2940 cubic inches = 40 gallons.
Second, run the truck at the truck stop until you have a 1/2 tank, or 20 gallons.
Third, add 10 gallons from the pump.
Fourth, insert the stick and mark it for 3/4 tank.
As Ray correctly pointed out, a 1/4 tank will be equidistant from the half tank mark as the three-quarter tank mark.
From Eliot in Estero, FL (just south of Cape Coral).
What is the matter U’s guys? the poor 18
wheeler guy, i am surprised he did drive off
the road. It is so simple. no need for
algebra… he knows the halfway point, right?
let us say it takes 400 gallons to fill it,
and the half ways point he drives 286 mile,
so a quarter of a tank will get him 143
miles. fill the tank, drive 143 miles, dip
the stick in, mark it. that would be 3
quarters full. drive 429 miles dip the stick
in, mark it that would be 1 quarter of tank
left.drive drive 71 miles you have used an
eight of a tank. and so on.
simple , hay
I have run into this several times here at our facility. We have several cylindrical (beer can) oil tanks that we monitor for levels using a dipstick. If Rick takes a yardstick and dips it into his tank the critical volume measurements are as follows:
0 inches = empty tank
6 inches = 1/4 tank
10 inches = 1/2 tank
14 inches = 3/4 tank
20 inches = full tank
He can then transfer these measurements to his dowel.
I’ve attached a spreadsheet that I use for this very purpose.
I hope this is helpful and that Rick has a sleeper cab just in case it doesn’t work.
Yet more advice from someone who has never driven a truck professionally.
Fuel economy varies by many factors, but on a semi, the biggest factor is the weight of the load, which varies widely. Miles driven while empty (deadheading) are minimized. One day a driver will be hauling 36,000 lbs. of freight, and the next, he might haul 10,000 lbs. of freight.
Hint: that’s not right. Should be more like .298 and .702 times the distance from full.
This is my story and I’m sticking to it.
Attached:notes, diagram, functioms
Let’s say the tank hold 50 gallons. When the tank is half full as determined by the measuring stick, add another 25 gallons, put in the stick again and you have the mark for 3/4. The 1/4 level mark would be at the same distance from 1/2 to 3/4’s but in the opposite direction.
Siphon the tank to empty. Fill the tank, keeping track of how much it holds. Siphon out 3/4ths of what it holds. Mark the dipstick accordingly.
I’m replying to myself. This is wrong. I didn’t visualize this correctly.
Consider a dipstick that’s long enough. Appropriatly mark a full line. Assume empty at the end of the stick (the part that goes into the tank). Mark the half point between full and end of stick. Call this point H. Measure distance from H to Full. Call this h. Mark distances plus and minus .2533h from H. These are 3/4 and 1/4 points on the stick.
I’m sticking to the half of the area of a semi-circle. The angle is 14.677 degrees, the sine of which is .2533 .
I’ll re-post later.
This is so easy, it’s not even funny, use a yard stick or two taped together if the tank is more than 3 feet. If the tank is 36" deep and the stick reads 18" when you stick it in you have half a tank, 9" a quarter and so on.
Alan Jacobs Junior College Graduate
I'm retracting post 141 because of an error in visualization. In case you're keeping score, the one quarter and three quarter full lines should be .2533 units from the half-way mark of the dip stick (not from FULL or Empty).
I reduced the problem to the consideration of the area of a semi-circle and the symatries that apply. I'm calling a FULL tank analagous to the area of the semi-circle equal to 1. We seek to find the conditions when thw area of the semi-circle is .5 (that level will indicate the level of the 3/4 full mark.
That level, being less than FULL involvrs considering the area of two sectors defined by 2 radii and an angle alpha plus the area of a triangle where the fuel level is the base opposite an angle of 2 times beta ( where 2 alpha + 2 beta = 180 degrees).
The area of the above region equal to .5 is satisfied when alpha is 14.677 degrees. The sine of alpha is .2533 and 1 minus the sine of alpha is .7467 .
Revised notes attached.Wrong, but works for rectangular fuel tanks.
Is that Alan Jacobs, Jr., college graduate -or- Alan Jacobs, junior college graduate?
This does not require integral calculus, but trigonometry. Imagine the cross-section of the tank as a circle with center c and a radius of 1. The horizontal diameter is acb. The vertical diameter is icj. Now imagine a horizontal chord, xyz. The top of the circle described by aib is half a tank. To find the next quarter of a tank, we need to set the position of xyz so that the area within abzx is equal to the area within xzj. This is equivalent to setting the area within acyx equal to 1/8th the area of the entire circle.
Call the angle acx angle v. acyx may be divided into two parts: a pie slice with angle v and radius 1. Its area is therefore v/2 (v being expressed in radians.) There is also a right triangle xcy, whose area is sin(v) x cos(v)/2. When v is set to 0.4158 radians, then the sum of these two areas is pi/8. That’s about 23.8 degrees. The sine of v is 0.4039.
When the fuel level is 0.702 of the way down, Rich has a quarter tank left.