Nice try my BUTT. You know I’m right! By the way… YOU have personally never provided any concrete calculation of the correct answer. YOU have only provided critic, nothing else. I have looked through all 290 plus comments for your absolute calculation… WHERE are your calculatuions? NOT TO BE FOUND ANYWHERE!! Critic but no real calculation.
Displacement of liquid is a Real “loss” that needs to be accounted for. YOU… I think are a person with an AS degree nothing more, a Designer from ITT Technical Institute!! I think we should have a competition to see who is really good at real life … real time engineering problems. I WILL WIN. I am R I G H T and you are wrong! PERIOD. The dip stick will cause the liquid level to R I S E… You know it … and so does everyone else.
I don’t recall if I posted it in this thread, or another on the subject, but I calculated it the same as some others did, using geometry and trigonometry. I explained it in a post (but perhaps it was on the same subject but a different thread). I wrote formulas in a spreadsheet to calculate the percentage of a circle under a chord at a given height, then just used the ?Solver? function to calculate the distance for 25%.
Displacement of liquid is a Real “loss” that needs to be accounted for? I think we should have a competition to see who is really good at real life … real time engineering problems. I WILL WIN. I am R I G H T and you are wrong! PERIOD. The dip stick will cause the liquid level to R I S E… You know it … and so does everyone else.
Try this one on for size, you engineering genius ? the problem was presented to us as a mathematical problem to calculate the area of a partial circle. To accurately account for the displacement caused by the measuring stick/dowel, you would need to know not only the dimensions of the stick, but the total volume of the tank, which we don?t know. Incidentally, a half-inch diameter dowel at a depth of 5.96 inches would displace about 5 thousandths of one gallon. The difference between your 6.1 inches and the correct answer of 5.96 is more than .6 gallons in a 75 gallon tank.
You can draw it in something like AutoCad and let it do the measuring, or use a tank simulation program like:
Either way, you?ll see that 6.1 inches is a little too high. But, of course, you realize this, which is why you started in on this nonsense about measuring stick displacement, even though your calculation doesn?t include anything to account for it, it?s just wrong.
It really doesn?t appear that you are good at real life engineering problems. Perhaps you should look into ITT Tech.
Wow, talk about overkill. It’s like reading Stephen Hawking theorys to explain quantum theory to kindergartners.
The truck most likely has a saddle pump system, so one tank can be empty. Drive one empty, trasnfer fuel to the other tank, whatever works, siphon off any remaining fuel in the tank to make sure it’s as empty as possible.
The tank has a known volume but for ease I’ll use 100 gallons.
Get a fairly thick dowel (like 1" in diameter) nothing that will cause a lot of displacement, so not a 2x4 lol) and a knife (No marker, fuel will remove marker). Put 10 gallons in the talk, slide in the dowel in for a second and make a notch at the fuel level on the dowel. Do this every 10 gallons, and also at the quarter marks (25, 50, and 75 gallons).
Make sure that the dowel is put in the same position each time, like pressed against the same point against the inside of the filler neck. Measure each mark and write them down on a few 3x5 cards (maybe laminate them) and that way if the dowel breaks he can just get a new dowel at any hardware store and mark a new dowel. Keep a card in the glove box, log book, one at home, add it to his iPhone if he wants to(is there an app for that?).
Looking over some other issues about tank curvature - With my method the curve of the tank isn’t going to be as much of a factor from 10 gal mark to 10 gal mark as it would be from quarter mark to quarter mark, so figuring a gallon or two, or five, would be pretty easy without a significant margin or error; and MPG some simple math.
Hi you guys,
This is Frank from Phillipsburg NJ.
OK,
1-first find an empty coffee can, one that have one of those plastic lids so you can see thrugh it.
2- Fill it 1/4 full and turn it sideways marking the level on the plastic lid.
do the same ting for 1/2 and 3/4.
3- Measure the distances from the bottom quad of the lid to each mark.
4- measure the diameter of the coffee can. (suppose 4")
5- divide the dia. of the can into the diameter of the actual fuel tank, 20"? (20 divided by 4 lets say equals 5)
6- multiply each measurment on the coffee can by 5 ( the multiplyer) and that will be the depth of each quarter on the actual fuel tank (Scaled Up)
Thanks Frank
With the paper make a little cylinder tank. Leave one end uncapped.
Measure the length tank.
Stand the tank up and fill it 1/4 of the way with salt.
Cap the open end of the tank.
Lay the tank on its side and measure up to the level of salt.
Now we have the percentage of the diameter of a cylindrical tank that is 1/4 full.
Use that percentage on the real fuel tank. Subtract that measurement from the top of the cylinder for 3/4, add it to the bottom for 1/4.
With the paper make a little cylinder tank. Leave one end uncapped.
Measure the length tank.
Stand the tank up and fill it 1/4 of the way with salt.
Cap the open end of the tank.
Lay the tank on its side and measure up to the level of salt.
Now we have the percentage of the diameter of a cylindrical tank that is 1/4 full.
Use that percentage on the real fuel tank. Subtract that percentage from the top of the cylinder for 3/4, add it to the bottom for 1/4.
I see a number of correct answers, but I will add mine.
The average value for the height of a semicircle is 2/pi or .681689(approximately) from the center line of the circle to the outer edge. The oil level in the rounded lower quarter of the horizontal cylinder would be 6.81689 inches (or there abouts) from the bottom for a tank with a 10inch radius.
The half full mark would be 10 inches, and the 3/4th mark would be radius x (2 - 2/pi) which is 13.18310 inches(+ or- a smidgen).
So I guess if I was trucken down the road at night and I wanted to measure the tank, 7", 10", and 13" would be good numbers to remember if I had 20in saddle tanks and a 24" dipstick.
MATH: The Mean Value and Average Value Theorem For Integrals http://ltcconline.net /greenl/courses/105/antiderivatives/secfund.htm
I now have three equations for segment area (one from the mathworld.wolfram website) all of equal validity, and all requiring successive approximations for specific answers. My equations are of the form A = f(d), where d is the radial distance. I am trying to write one in the form d = f(A) with no success. Any help out there?
Just heard the “solution” on air and sent in an email. Balancing a semicircle doesn’t work. Balance is based on area times distance. If you divide this way you have two areas. The area of the lower half is concentrated closer to the balance point, so the distance to the CG of the lower half is shorter. To get balance the lower half has to have larger area.
An ugly method would be to use the cardboard box to make a balance and then by trial and error put equal areas of cardboard on the two measure points.
The Jan 22 show repeated this question, but the balance point answer is incorrect. This is now being discussed in a couple of new Topics, which might be fresh and concentrate on this “balance” issue only.