Formula for Diameter of Cylinder Using Height/Volume.

  1. I need the formula for finding the Diameter of a Cylinder using it's Volume and Height.

    Thanks!

    Here is the problem. Find Diameter of Cylinder.
    Volume = 1256.64
    Height = 16 Inches
    Diameter = ?

    On my own...

    Knowing Volume = (Pi x Diameter Squared x Height) Divide by 4

    I came up with this forumla for D?

    D = Square Root of (4V/H/Pi)

    Plugging in the numbers.... D = 10 Inches?

    Is there an easier forumla?
     
    Last edited: Mar 22, 2008
  2. jcsd
  3. Gib Z

    Gib Z 3,348
    Homework Helper

    Well, what is the Cylinders volume is given by the product of the base circles area and its height. What is the circles area given by, in terms of the radius? How are the radius and diameter related?

    PS: Welcome to Physics forums!! =]
     
  4. Gib Z

    Gib Z 3,348
    Homework Helper

    Well you that right, [tex]D= \sqrt{ \frac{4V}{h\pi}}[/tex]. Easiest formula there is lol. Would look nicer in terms of the radius though.
     
  5. #38

    #38 2

    I know this thread is old but it is the only one I could find that is most relevant to my problem.

    Can you explain it in terms of the radius?

    I am trying to figure out the same type of problem and am horrid with roots.
    I need to find the diameter given H=10m and V=125.6m cubed
     
  6. #38

    #38 2

    Here is the formula I used to figured it but I think it is wrong.
    (I don't know how to do all the fancy stuff so I will just type it out.)

    v/h over Pi=r squared

    Edit: I think I got it figured now

    New formula I am using is v/h*Pi= r squared
    Find the square root of r squared then multiply * 2 for the diameter
     
    Last edited: Mar 7, 2010
  7. HallsofIvy

    HallsofIvy 40,539
    Staff Emeritus
    Science Advisor

    The volume of a cylinder of height h and radius r is [itex]V= \pi r^2h[/itex].
    To solve that for the radius, divide both sides by [itex]\pi h[/itex] and take the square root:
    [tex]r= \sqrt{\frac{V}{\pi h}[/itex]

    The diameter is twice the radius, of course, so
    [tex]d= 2\sqrt{\frac{V}{\pi h}= \sqrt{4V}{\pi h}[/itex]
    because [itex]\sqrt{4}= 2[/itex].
     
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