# Car washing

Discussion in 'Physics & Math' started by Pim, May 27, 2003.

1. ### everneoRe-searcherRegistered Senior Member

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Pim,
if i say the width is 1.231m approx., you should not be surprised.

3. ### PimRegistered Senior Member

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everneo: that is indeed the correct answer. Would you care to explain how you obtained it?

5. ### everneoRe-searcherRegistered Senior Member

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2,621
the link to the answer is also available in that page. sorry i could not overcome the temptation..

*ducks and runs away*

7. ### everneoRe-searcherRegistered Senior Member

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*deleted*

Last edited: Jul 9, 2003
8. ### PimRegistered Senior Member

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I don't understand your last post?

9. ### everneoRe-searcherRegistered Senior Member

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me too. i'm going to delete my 'last post'.

10. ### PimRegistered Senior Member

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Well, as everneo pointed out the answer is on the same page as the picture..(thnx for that). There is also an explanation of how it is derived. It is in Dutch though, good luck

11. ### PimRegistered Senior Member

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Well, i hope people will try to solve it without looking there

12. ### everneoRe-searcherRegistered Senior Member

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you are right. let me edit it again.

13. ### AbsaneRocket SurgeonValued Senior Member

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8,989

How did I do it?

First, I set up the distance forumla for both lines.
Then, I made a linear equation for both.
I knew f(x) = g(x) = 1, so I solved for 'x'.
I set both x's equal to eachother, and solved for 'w,' the width.

As far as anyone wanting to see the work, there is not much to it. Like I said in a previous thread, I visualize most stuff...

With all the work, you land to this:

1/sqrt(9-w^2) + 1/sqrt(4-w^2) - 1 = 0

James Sibley

Last edited: Jul 9, 2003
14. ### FunkyTurtle2Registered Member

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Back to the desert! The Car Problem(tm):

Geez, this has taken me forever. Using calculus and separating into variables, dg/dx (change in gas per unit distance) = - (amount of gas needing to be transfered) / 20, the transfer size, * the MPG. The rate is defined to be negative. I am transfering the whole load as I go, shuttling it back and forth a meter at a time, so the number of trips I take is almost entirely dependent upon how much gas I have, not how far I am traveling. Then

200/G dg = - dx

200 ln g = - x + C, a constant

using the point (g, x) = (20, 300), having ... is that where I screwed up. Well, the latest place. (I was off by a factor of 10, couldn't figure out why, etc. etc., and now I see I did it in another place too.) 20, 300, because once you get there you can just take off and can stop worrying about this back-and-forth crap. Anyway, you get

200 (ln 20) + 300 = C

Solving for g from the previous equation,

g = e^([C - x] / 200)

x = 0 should give the original amount of gas before setting out with my tons of fuel tanks.

g = e^( [200 (ln 20) + 300] / 200 - 0 )
= e^( ln 200 + 300/200 ), where the ratio 300/200 is derived from the "final" location of 300 miles and the product of 20 gallons * mpg.

g = 20 e^1.5

Which is, presumably, incorrect. (It's about 89 gallons.) Can anyone tell me what I did wrong?

If you say that you can only leave 10 gallons at a time, your fuel load is effectively 10 gallons (the unused 9.99 gallons just stay in your tank), and you get g = 20 e^(3/1) = 401.

Just to doublecheck -- this was when I was still getting "reasonable" values of about 430 and 460 gallons for the 20/10 tank size calculations, I tried writing a program for it.

I based it off of Persol's sequence, so I'd get 20, 35, 65... gallons needed at 300, 250, 200... miles respectively, working backwards. Here is the relevant code, after literally hours of finding at isolating bugs:

Code:
    Const curMileage As Currency = 10
Const curLoad As Currency = 20
Dim curMyLocation As Currency
Dim curStepsize As Currency
Dim curMyGas As Currency

curStepsize = Val(txtInputStep.Text)
curMyGas = 20
curMyLocation = 300
Do Until curMyLocation <= 0
curMyGas = curMyGas + DeltaGas(curMyGas, curStepsize)
curMyLocation = curMyLocation - curStepsize
Loop

Private Function DeltaGas(curMyGas As Currency, curStepsize As Currency)
'calculate gas needed to iterate the step

Dim curDelta As Currency

'stepsize / mileage = gas per trip
'tip: next time, dimension #of trips as a separate variable!

If Round(curMyGas / (curLoad - curStepsize / curMileage)) < curMyGas / (curLoad - curStepsize / curMileage) Then
curDelta = curStepsize / curMileage * (2 * (Round(curMyGas / (curLoad _
- 2 * curStepsize / curMileage) + 1)) - 1)                                      '2 * single-trip = roundtrip for TotalLoad/perload,
'then subtract one because don't go back.

ElseIf Round(curMyGas / (curLoad - curStepsize / curMileage)) > curMyGas / (curLoad - curStepsize / curMileage) Then
curDelta = curStepsize / curMileage * (2 * (Round(curMyGas / (curLoad _
- 2 * curStepsize / curMileage))) - 1) 'rounding is STUPID in VB! have to do two cases since I don't know the language very well.

Else
curDelta = curStepsize / curMileage * (2 * curMyGas / (curLoad _
- 2 * curStepsize / curMileage) - 1)
End If
DeltaGas = curDelta
End Function

It's in Visual Basic -- sorry, hehe, but it's the only programming language I am learning. btw, I managed to get Persol's 965 value somewhere along the way, but that was without rounding up to the nearest whole number of one-way trips... grrr. Something like that. It's why I had to code the stupid rounding bits in the code.

Anyway, some sample values are:
Stepsize (distance between stations) | gallons required at start:
0.5 | 422
1 | 425
5 | 462
10 | 508
20 | 735
25 | 566.67 (has to do with the non-linear nature of it. how many refueling trips to make, how many total stations, etc. One more station can make a significant difference.)
30 | 740
33 | 1019
40 | 1532
49 | 2348
50 | 1240
59 | 4174
60 | 2150
75 | 4385
90 | 190910
98 | 123,774,510
99 |1,990,297,029.8

Kind of surprising, since the peak of distance * fuel deposited does peak at 50 miles. I will not guarantee these values are accurate, but until someone finds a hole in my logic or coding ineptitude I will view them as precedent. Too bad my calculus is faulty... gr.

peace. :m:

15. ### phoenix2634Registered Senior Member

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86 drums of fuel

It takes 86 drums to cross the desert.

3 drums to place a drum at 100 miles.
15 drums to place a drum at 200 miles.
63 drums to place a drum at 300 miles.

3 + 15 + 63 + 3(drums at each station) + 2 (drums from starting point) = 86 drums of fuel.

This follows the method that ryans posted earlier.

The problem with the formula to get 42 drums of fuel is the distance unit. The maximum distance you can travel to allow you to shuttle fuel between stations is 100 miles, not the 200 miles that was used.

Phoenix.

Last edited: Jul 13, 2003
16. ### FunkyTurtle2Registered Member

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You are right; the original equation should have been dG = (2 * dx) (G/20) [<=the number of trips taken, no units] * 1 gallon / 10 mi.

Unit analysis is miles * gallons / miles.

And that is assuming that the amount of gas spent in shuttling fuel around is negligible when compared to how much there is in all, since in reality less trips would have to be made.

That still gives an end equation of 20 ln (e^3), or the 401 gallons from the "conservative" estimate. This is still different from the one from the program, lower than the answer on the website, and much lower than anything anyone else was able to come up with!

so what am I doing wrong?

17. ### PimRegistered Senior Member

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64
This is another problem i have been thinking about for a while.

Let a coordinate system be given. A man is standing in point (0,0). His dog is standing in (0,2). The man starts walking along the positive x-axis with constant speed. The dog starts walking in the direction of the man with the same speed. The dog hence walks along a curved line. The question now is if an explicit expression can be determined for this curved line, that is, a function y in terms of x.
What do you think?

Last edited: Jul 14, 2003
18. ### AbsaneRocket SurgeonValued Senior Member

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8,989
No response to my method... lol. Oh well. I am curious how everyone else did it. I would like to see other methods.

As for the man/dog problem... I am thinking about that one. I got something close...

something like f(x) = sqrt[ (r/r_1)^2 - 1 ]*x + 2

I just have to get rid of r/r_1 and convert it to something involving x...

JS

19. ### PimRegistered Senior Member

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4DHyperCubix: You mean your solution to the ladder problem? I think it's a very elegant way to do it like that

I used geometrical properties of the triangles in the picture to compute the width. Don't feel like posting that here though

20. ### AbsaneRocket SurgeonValued Senior Member

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What kind of geometrical properties? Knowing that will be good enough for me

How far along have you got with that man/dog problem?

21. ### PimRegistered Senior Member

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Well, i don't know the english word for it..i think it's something like congruent triangles? Does that makes sense? ("Gelijkvormig" in Dutch)

I haven't been able to come any further with the dog-man problem. I tried to determine the equation for the derivative of the curve the dog travels, but that is not that easy since the dog moves too...

22. ### MacMRegistered Senior Member

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10,104
Thick after coffee

Gentlemen,

I guess I am just thick even after coffee. Can sombody explain to me why we are talking about a curve in the dog/man problem?

It seems to me that the line created is a linear or straight line since both have the same speed. It matters not what direction and for that matter that they have the same speed.

Plotting coordinates of motion of two uniformly moving bodies seems to result in a straight line.