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Simple dipstick problem

  • Thread starter pat666
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  • #26
Mentallic
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Sorry about the half-assed response before, I had to run.

No no, it is just [tex]\frac{H}{h}=\frac{R}{r}[/tex] because the ratio between the big cone radius (R) to the little cone radius (r) is the same as the big cone height (H) to the little cone height (h). So you pretty much have that problem answered once you solve for r in terms of the other known constants and then plug that into the volume of the small cone formula.

I'm sure the formula for the cylinder can be a lot simpler than that, I'll solve the problem now and then get back to you.
 
  • #27
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No worrys, You've been way more than helpful so far. THANKS
r=(h/H)*R

[tex] V=(\pi*R^2*h)/3 -\pi*((h/H)*R)^2(H-h)/3 [/tex] hope I've done the LaTex correctly.

The way I read that is V entire cone minus V water filled portion of cone and what that yields is the volume of the empty portion of the cone. So my question is should That answer now be subtracted from the volume of the cone again??

Thanks
 
  • #28
Mentallic
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so the Area of the liquid filled portion is A = pi*r^2/2 - r^2*arcsin(1-h/r) - (r-h)*sqrt(h(2r-h))
Can you show me how you got that, because I'm getting something else.
 
  • #29
Mentallic
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No worrys, You've been way more than helpful so far. THANKS
r=(h/H)*R

[tex] V=(\pi*R^2*h)/3 -\pi*((h/H)*R)^2(H-h)/3 [/tex] hope I've done the LaTex correctly.
Don't put spaces in the [tex ] and [/tex ] tags and then you'll get latex code :wink:
edit: never mind, you fixed it already.

The way I read that is V entire cone minus V water filled portion of cone and what that yields is the volume of the empty portion of the cone. So my question is should That answer now be subtracted from the volume of the cone again??

Thanks
Sure, you can do it that way, but why not take the much easier approach? You already know the volume of the smaller water-filled cone is [tex]V=\frac{\pi r^2 h}{3}[/tex] and you don't know r, but you have already shown what r is in terms of R, H and h. Just plug it into there and you're set. You should also get the same answer from doing it the longer way as you've suggested.
 
  • #30
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Theres every chance that I'm wrong, as I said I am very rusty at this. Perhaps this picture will help illustrate what I'm thinking. As I'm looking at the pic I just drew I think I may have stuffed something up, its not making sense to me now.
 

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  • #31
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oh yeah, don't know why I missed that.
so [tex] V_(fluid)=(\pi((h/H)*R)^2*h)/3 [/tex] probably simplify down further.

Thanks
 
  • #32
Mentallic
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Theres every chance that I'm wrong
Haha fair enough :biggrin:

Since we've already failed a few times at understanding each other with much simpler diagrams, I'll have to draw up a picture for this more complicated one.

First of all, we can save ourselves a lot of time if we don't have to prove the area of the segment of a circle. The wiki article for it was already posted:
I found this on wikipedia... which might help http://en.wikipedia.org/wiki/Circular_segment" [Broken]
If you don't understand it though just ask and I'll walk you through it.

Ok so we have that the area of the segment of the circle is [tex]A=\frac{r^2}{2}\left(\theta-\sin\theta\right)[/tex]

But we don't know [itex]\theta[/itex] so again we're going to have to do something like in the other question to find [itex]\theta[/itex] in terms of other constants that we do know, mainly h (the height of the water level) and r (the radius of the circle).

[PLAIN]http://img17.imageshack.us/img17/1326/circleo.png [Broken]

The angle [tex]\theta[/tex] sweeps the entire brown angle in the circle picture, and therefore due to symmetry, the brown angle in the triangle picture is going to be [tex]\frac{\theta}{2}[/tex]

Now you can easily get your relationship and plug it back into the equation. And by the way, [tex]\sin\left(\cos^{-1}\left(x\right)\right)=\sqrt{1-x^2}[/tex]
 
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  • #33
Mentallic
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[tex] V_(fluid)=(\pi((h/H)*R)^2*h)/3 [/tex] probably simplify down further.
Sure, [tex]V=\frac{\pi h^3R}{3H^2}[/tex]

By the way, you can click on my LaTeX to see how it is written, in case you're interested.
 
  • #34
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I get the circular segment part (to some degree), what I don't get is that the adjacent side of your triangle is labeled r-h, I can't see how that can be true?? wouldn't r=h and therefore your triangle be non existent or is this a misprint or an error on my part? Also just to confirm, we are simply trying to get theta in terms of the depth of the water?

P.S just to confirm the last post I made on the cone question is correct?
Thanks
 
  • #35
Mentallic
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what I don't get is that the adjacent side of your triangle is labeled r-h, I can't see how that can be true?? wouldn't r=h and therefore your triangle be non existent
Why does r need to equal h? r is the radius of the circle and h is the height of the water. First let [itex]h<r[/itex], so that the side of the triangle is r-h because it is what is left of the radius after taking away h. For r=h we have a degenerate case where the water fills up half the circle, so it would simply be [tex]A=\frac{\pi r^2}{2}[/tex]. You can also take [itex]h> r[/itex] and try find the formula for that, but it'll be the same as for [itex]h<r[/itex].

Also just to confirm, we are simply trying to get theta in terms of the depth of the water?
In terms of the depth of the water AND the radius of the circle. We can't do it in terms of the depth of the water alone. Imagine we had a circle radius 1 and the depth of the water was very close to it, about 0.99. The angle [itex]\theta[/itex] will be very close to [itex]\pi [/itex]. Now what if we kept the water depth the same, but made the circle very big now? Well the angle will get a lot smaller. So obviously [tex]\theta[/tex] is dependent on more than just the depth of the water.

P.S just to confirm the last post I made on the cone question is correct?
Thanks
Well you tell me, what makes you unsure if it is correct or not?
 
  • #36
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I think it is correct but given the amount of trouble I've had getting the solution you can see why I am unsure. just the way the picture was drawn made me think r=h, which would be true if the water was at the centre. I get that [tex] \theta=2cos^-^1(r-h)/r [/tex]
not sure if thats correct because you gave me some trig info that was a bit more complex.
also this will only work to the halfway point but I was thinking I would just do the reflection of the earlier dipstick points for points after the mid line.
thanks
 
  • #37
Mentallic
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I think it is correct but given the amount of trouble I've had getting the solution you can see why I am unsure.
Yes it's correct. I was just showing you that you can tidy up
[tex] V_(fluid)=(\pi((h/H)*R)^2*h)/3 [/tex] probably simplify down further.
into
Sure, [tex]V=\frac{\pi h^3R}{3H^2}[/tex]

just the way the picture was drawn made me think r=h,
Oh, yeah that would be my fault, sorry :smile:

which would be true if the water was at the centre.
The blue line was meant to be where the water level was at.

I get that [tex] \theta=2cos^-^1(r-h)/r [/tex]
Yes that's right.

not sure if thats correct because you gave me some trig info that was a bit more complex.
My trig info was wrong, ignore it. I forgot about the 2 that was going to be in front of it. What I was meant to give you was
[tex]\sin\left(2\cos^{-1}\left(x\right)\right)=2x\sqrt{1-x^2}[/tex]
It may look complex, but its purpose is simple. When you plug [itex]\theta[/itex] into the area equation [tex]A=\frac{r^2}{2}\left(\theta-\sin\theta\right)[/tex] you're going to be left with
edit: [tex]\sin\left(2\cos^{-1}\left(\frac{r-h}{r}\right)\right)[/tex] which is where you can simplify this with the equality I gave above.
You already have the answer, but it's just if you wanted to simplify things a bit more.

Markers wouldn't give full marks if you left an answer as [tex]\sin\left(\sin^{-1}\left(x\right)\right)[/tex] so I doubt they would give full marks if you left it as [tex]\sin\left(\cos^{-1}\left(x\right)\right)[/tex] either.

also this will only work to the halfway point but I was thinking I would just do the reflection of the earlier dipstick points for points after the mid line.
thanks
I believe the same formula will work for the water level anywhere from 0 to 2r (the diameter of the circle) but I'll check to see.
 
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  • #38
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Ok thanks alot for all your help
 
  • #39
OmCheeto
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The cylinder on it's side problem reduces to y=x - sin(x) in it's simplest form.
If you can solve for x, then you can solve the problem.
There is a reason they put this problem in computer science text books and never in mathematics text books.
 
  • #40
Mentallic
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Ok thanks alot for all your help
No worries

The cylinder on it's side problem reduces to y=x - sin(x) in it's simplest form.
If you can solve for x, then you can solve the problem.
Can you please elaborate?

There is a reason they put this problem in computer science text books and never in mathematics text books.
I've seen questions similar to this in maths books.
 
  • #41
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This question is for yr 11 math. My solution is I believe [tex] V=L*\sin\left(\cos^{-1}\left(2\cdot\frac{r-h}{r}\right)\right) [/tex].
 
  • #42
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This question is for yr 11 math. My solution is I believe [tex] V=L*\sin\left(\cos^{-1}\left(2\cdot\frac{r-h}{r}\right)\right) [/tex].
That's not right. The formula is [tex]A=\frac{r^2}{2}\left(\theta-\sin\theta\right)[/tex] where [tex]\theta=2\cos^{-1}\left(\frac{r-h}{r}\right)[/tex]
 
  • #43
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whoops I forgot the first theta, so V=L*(r^2/2(2arccos(r-h/r)-sin(2arccos(r-h/r))) pretty messy.
 
  • #44
OmCheeto
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Can you please elaborate?
Perhaps the question in the book was worded differently.
All I know is that if you have a 10 gallon tank, it is not possible to place integer gallon marks on the dipstick. Unless of course you are clever enough to solve for x.


And perhaps I should give some https://www.physicsforums.com/showthread.php?p=3042826#post3042826" regarding this particular problem before someone yells at me for playing mind games.
 
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