PRO: A very interesting trick. I think your drawing and logic would be clearer if you shaded the areas A and C.
CON: I am not sure that estimating the equality of areas A and C is any easier than just estimating the 1/4 volume line in the original circle.
Was that Peter? or Bill?
Anyway, you got it.
For rhombus (semi rhombus = triangle) that’s the answer:
“square root of one-half” measured as you explained.
And about the impossibility of splitting the semicircle:
as we already know, it can be done, just not in a “nice” way.
Instead of “impossible” we should say that halving the semicircle is “ugly” or “messy”.
By comparison, the triangle/rhombus case is “nice” or “clean”.
Keep up the good work.
Cheers,
–Ion
I don’t quite know why that would supposedly work anyway (the centroid answer). I plotted that in AutoCAD to let it calculate the area. It’s easy to calculate the centroid and area of any shape (great for estimating areas from aerial pictures!). The calculation of the centroid was correct, but that’s definitely not the quarter tank mark!
Draw circle 1 with diameter.
Draw circle 2 with same diameter as circle 1.
Move circle 2 up and down untill area A is equal to area C.
Draw chord shown
At this point area under chord is 25% of total area.
Measure X ( about 1.25)
Measure Y (About 4 )
Height on stick = 1.24 / 4 or 31% of diameter of tank.
(My other posting had an error.)
if you estimate 1/4 of the area of the circle the error will be much greater.
I did not see what I think is the most straightforward (mathematical) solution:
If you draw 2 radii
from the center of a circle to the surface
of the liquid, the angle between them is t.
The cross section of the liquid is a
segment of a circle. The area of a segment
is A=(r^2/2)(t-sin(t)). For 1/4 tank, the
area must be A=(pir^2)/4. The result is
f(t)=0=2t-2sin(t)-pi. This is a
transcendental equation. (It cannot be solved for t.)
The Newton-Ralphson method can be used to iterate for
the solution. Use an initial guess of pi/2.
Your next guess is
(pi/2)-(f(pi/2))/f’(pi/2)). Four iterations shows
t=2.30988 radians. The depth of the fuel is
then found from d=r(1-cos(t/2)). The final
result is that the depth must be about 60%
of the radius of the tank (r*0.596027).
OK you have too many wrong answeres here so I did it with a spreadsheet on a quarter circle of unit radius As you can see at .41 the area is 50.65 percent. .41 is .69 from the circle edge and a unit circle has a diameter of 2 so .69 is 34.5 percent of the diameter. Sorry about formatting.
zero to 1 root(1- squ) sm area tot area percent
0.01 0.99994999875 0.00999949999 0.00999949999 1.2733350296
0.02 0.99979998 0.0099979998 0.01999749979 2.5464790255
0.03 0.9995498987 0.00999549899 0.02999299877 3.8193045683
0.04 0.99919967974 0.0099919968 0.03998499557 5.0916841426
0.05 0.99874921777 0.00998749218 0.04997248775 6.3634900993
0.06 0.99819837708 0.00998198377 0.05995447152 7.6345946161
0.07 0.99754699137 0.00997546991 0.06992994143 8.9048696593
0.08 0.99679486355 0.00996794864 0.07989789007 10.174186944
0.09 0.99594176537 0.00995941765 0.08985730772 11.442417894
0.1 0.99498743711 0.00994987437 0.09980718209 12.709433604
0.11 0.99393158718 0.00993931587 0.10974649797 13.975104796
0.12 0.99277389168 0.00992773892 0.11967423688 15.239301781
0.13 0.99151399385 0.00991513994 0.12958937682 16.501894413
0.14 0.99015150356 0.00990151504 0.13949089186 17.762752051
0.15 0.98868599666 0.00988685997 0.14937775182 19.021743515
0.16 0.98711701434 0.00987117014 0.15924892197 20.278737039
0.17 0.98544406234 0.00985444062 0.16910336259 21.533600228
0.18 0.98366661019 0.0098366661 0.17894002869 22.786200012
0.19 0.98178409032 0.0098178409 0.1887578696 24.036402597
0.2 0.97979589711 0.00979795897 0.19855582857 25.28407342
0.21 0.9777013859 0.00977701386 0.20833284243 26.529077095
0.22 0.97549987186 0.00975499872 0.21808784114 27.771277365
0.23 0.97319062881 0.00973190629 0.22781974743 29.010537047
0.24 0.97077288796 0.00970772888 0.23752747631 30.246717982
0.25 0.96824583655 0.00968245837 0.24720993468 31.479680973
0.26 0.96560861637 0.00965608616 0.25686602084 32.70928573
0.27 0.96286032217 0.00962860322 0.26649462406 33.935390814
0.28 0.96 0.0096 0.27609462406 35.157853567
0.29 0.9570266454 0.00957026645 0.28566489052 36.376530054
0.3 0.95393920142 0.00953939201 0.29520428253 37.591274994
0.31 0.95073655657 0.00950736557 0.3047116481 38.801941691
0.32 0.94741754259 0.00947417543 0.31418582352 40.008381959
0.33 0.94398093201 0.00943980932 0.32362563284 41.210446052
0.34 0.94042543564 0.00940425436 0.3330298872 42.40798258
0.35 0.93674969976 0.009367497 0.3423973842 43.600838431
0.36 0.93295230318 0.00932952303 0.35172690723 44.788858682
0.37 0.92903175403 0.00929031754 0.36101722477 45.971886511
0.38 0.92498648639 0.00924986486 0.37026708963 47.149763101
0.39 0.92081485653 0.00920814857 0.3794752382 48.322327543
0.4 0.91651513899 0.00916515139 0.38864038959 49.489416731
0.41 0.91208552231 0.00912085522 0.39776124481 50.65086525
0.42 0.90752410436 0.00907524104 0.40683648585 51.806505266
0.43 0.90282888744 0.00902828887 0.41586477473 52.956166399
0.44 0.89799777283 0.00897997773 0.42484475246 54.099675596
0.45 0.89302855497 0.00893028555 0.43377503801 55.236856998
0.46 0.88791891522 0.00887918915 0.44265422716 56.367531792
0.47 0.8826664149 0.00882666415 0.45148089131 57.491518058
0.48 0.87726848798 0.00877268488 0.46025357619 58.608630611
0.49 0.87172243289 0.00871722433 0.46897080052 59.718680825
0.5 0.86602540378 0.00866025404 0.47763105455 60.821476449
0.51 0.86017440092 0.00860174401 0.48623279856 61.916821414
0.52 0.85416626016 0.0085416626 0.49477446117 63.00451562
0.53 0.84799764151 0.00847997642 0.50325443758 64.084354715
0.54 0.8416650165 0.00841665017 0.51167108775 65.156129854
0.55 0.83516465442 0.00835164654 0.52002273429 66.21962744
0.56 0.82849260709 0.00828492607 0.52830766036 67.27462885
0.57 0.82164469207 0.00821644692 0.53652410728 68.320910134
0.58 0.81461647418 0.00814616474 0.54467027202 69.358241694
0.59 0.80740324498 0.00807403245 0.55274430447 70.386387937
0.6 0.8 0.008 0.56074430447 71.405106898
0.61 0.79240141343 0.00792401413 0.56866831861 72.414149829
0.62 0.78460180984 0.0078460181 0.57651433671 73.413260755
0.63 0.77659513261 0.00776595133 0.58428028803 74.402175988
0.64 0.76837490849 0.00768374908 0.59196403712 75.380623598
0.65 0.75993420768 0.00759934208 0.59956337919 76.348322831
0.66 0.75126559884 0.00751265599 0.60707603518 77.304983469
0.67 0.74236109812 0.00742361098 0.61449964616 78.250305127
0.68 0.73321211119 0.00733212111 0.62183176727 79.183976477
0.69 0.72380936717 0.00723809367 0.62906986095 80.105674385
0.7 0.71414284285 0.00714142843 0.63621128938 81.015062954
0.71 0.70420167566 0.00704201676 0.64325330613 81.911792453
0.72 0.69397406292 0.00693974063 0.65019304676 82.795498123
0.73 0.68344714499 0.00683447145 0.65702751821 83.66579883
0.74 0.67260686883 0.00672606869 0.6637535869 84.522295543
0.75 0.66143782777 0.00661437828 0.67036796518 85.364569614
0.76 0.64992307237 0.00649923072 0.6768671959 86.19218081
0.77 0.63804388564 0.00638043886 0.68324763476 87.004665065
0.78 0.62577951389 0.00625779514 0.6895054299 87.801531885
0.79 0.61310684224 0.00613106842 0.69563649832 88.582261342
0.8 0.6 0.006 0.70163649832 89.346300563
0.81 0.58642987646 0.00586429876 0.70750079708 90.093059606
0.82 0.57236352085 0.00572363521 0.71322443229 90.82190657
0.83 0.5577633907 0.00557763391 0.7188020662 91.532161747
0.84 0.54258639865 0.00542586399 0.72422793018 92.223090562
0.85 0.52678268764 0.00526782688 0.72949575706 92.893894952
0.86 0.51029403289 0.00510294033 0.73459869739 93.543702711
0.87 0.49305172142 0.00493051721 0.7395292146 94.171554133
0.88 0.47497368348 0.00474973683 0.74427895144 94.776385004
0.89 0.45596052461 0.00455960525 0.74883855669 95.357004544
0.9 0.43588989435 0.00435889894 0.75319745563 95.912066169
0.91 0.41460824883 0.00414608249 0.75734353812 96.440027775
0.92 0.39191835885 0.00391918359 0.76126272171 96.939096104
0.93 0.36755951899 0.00367559519 0.7649383169 97.407145918
0.94 0.34117444218 0.00341174442 0.76835006132 97.84159701
0.95 0.31224989992 0.003122499 0.77147256032 98.239215627
0.96 0.28 0.0028 0.77427256032 98.595767263
0.97 0.24310491562 0.00243104916 0.77670360947 98.905336747
0.98 0.19899748742 0.00198997487 0.77869358435 99.158739889
0.99 0.1410673598 0.0014106736 0.78010425794 99.338374882
1 0 0 0.78010425794 99.338374882
OK you have too many wrong answeres here...So you thought you'd add your incorrect answer to the mix? This looks like a bunch of nonsense, but in looking at your previous answers you posted, it looks like you're saying that the quarter tank mark is at 31.479680973% of the diameter. Attached is an AutoCAD representation of that portion of the circle. It is incorrect. Rather than using the trigonometry that many have used to calculate the correct answer (5.96 inches, approx 29.8% of the diameter), I used AutoCAD to draw what you calculated and let it calculate the area of the portion of the circle. And drawing it with the correct height (5.96 inches) gives you the correct quarter area of the circle.
T’was Peter.
I claim no great expertise but I have occasion to put a wrench on a few semi trucks.I think you guys are thinking way to much, that could be an over statement.This is what I did, I walked out side the shop rolled a 90 gal. fuel tank got out my abacus divided 90 gal.by 4 and got 22.5 gal. filled 90 gal tank with 22.5 gal then dipped my stick marked as 1/4 tank.If you need half tank, fill tank another 22.5 gal dip stick for 1/2 tank. so does this mean I get more work done because I use an abacus.
I hate to put a fly in the ointment, but you are assuming the tank will be completely full and that you are measuring from the top center of the tank. Most of the tanks I have seen have a slightly offset filler neck, so the tank never gets completely full because fuel will run out of the filler neck before it reaches the top.
So… your dipstick would actually enter from a slight angle and getting a true reading of the level of fuel in the tank becomes mathematically more complicated.
I’m no mathematician, but I do know that when you put the dipstick in from an angle, the reading will not be true. The half tank mark will not be 1/2 the diameter of the tank, it will be just below.
It seems to me you would need to know not only the angle off center of the fill opening, but the diameter of the opening as well so you would know at what point the fuel would spill from the tank, making it full.
So… how do you solve the problem now?
Let me let loose the fly from the ointment. I’m going to assume that you didn’t listen to the call, because Tom and Ray asked what the angle the stick entered the tank was - Rich’s response, “It’s vertical.” He stated that when completely full, the fuel came to 20 inches on the stick, he also stated that the diameter of the tank was 20 inches. So, not only is the stick going in the tank vertically, it must also be going in at the very top of the tank.
I imagine if the filler was offset and the stick was going in at an angle, and we knew the offset and angle, we could graphically represent the cross section of the tank and use a little trig to figure out the appropriate measurements (that’s how I’d try it anyway), but it’s almost 2 a.m. where I am, and I’m not going to try to figure it out tonight! I’m kind of curious though, and I might play around with it tomorrow.
OK… I did not hear the call… just thinking out loud…
Thank you for the correction.
Out of curiosity, I did this little drawing to see how an offset filler and angled stick would change the measurements. As you can see, with an offset filler, you wouldn?t be able to fill it up all the way. I just drew the ?full? line in their arbitrarily. In real life you would probably fill it past that. At the line I drew though, you would get about 90.5% of the tank capacity (the semis my father has on his farm have 100 gallon tanks and the full mark is at 93 gallons). So, the area under that line is .95100pi. I left the half and quarter tank marks at half the total capacity of the tank, not the total fillable amount. I imagine (and someone will surely correct me if I?m wrong) that if we wanted to find the half and quarter marks of the fillable amount, we would just recalculate the heights and instead of find the point where the area under the chord is .25100pi, we would find the vertical distance where the area is .905*.25100pi for a quarter tank, and .905.5*100pi for a half tank.
I think you could figure out these measurements using the angles and known distances and a little trig, I just let AutoCAD do it for me to save some time.
My friend, Sal, has already written a response which was not highlighted and I am disappointed in this. All the ideas I saw listed were computational and highly mathematical. His (Sal’s) idea was to drain out the tank, and then fill it to the top. Then drain out 3/4 of it and see where the level is. It is simple, easy and elegant - stuff that you guys seem to like often. Why not this time?
Jon
Its very simple. Either trucker Rich gets his PHD in math or uses his dowel to measure his gas consumption.
Whats so complicated about getting the gas gauge/sending unit fixed?
here is the solution using mathematica
Since the volume formula for both cones and pyramids is 1/3 * base * h, and while in one half or another, the area of the base increases proportionately to the square of the distance from an apex, the volume is a cubic function of the distance from the apex.
So in this problem, the volume must be one-half of the volume of the full pyramid or cone, and so the height from the base is (1 - (1/2)^1/3), or one minus about 0.79370052. This comes out to be .20629…
I would concur with the “messy”/“clean” analysis. Only the triangle solution can be constructed with a compass and straightedge.
Just fill the the tank up after every chicken fried steak