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    Head math

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    • Member since
      February, 2008
    Head math
    Posted by machz975 on Sunday, August 14, 2016 1:24 PM

    Can some body tell me how to figure out the percentage of the squish band width in a   Math formula

    Moderator
    • Member since
      October, 2013
    Posted by Ross Halvorson on Thursday, August 18, 2016 10:33 AM

    This article might be helpful for what you're looking for.

    http://www.amsnow.com/how-to-tech/2014/11/pressure-the-competition

    • Member since
      February, 2008
    Posted by machz975 on Saturday, August 20, 2016 1:22 PM

    Sorry this did not answer my question ,, but  good info on there

    mac
    • Member since
      September, 2006
    • From: New Jersey
    Posted by mac on Monday, August 22, 2016 10:12 AM

    Inline image 1

    In the example above i assumed a bore size of 3.5 and dome size of 2.75  so insert your own values.  Squish is the differance of dome and bore.

    The equation might seem like it's missing a factor of Pi, but if you work through the calculations, you'll find that it cancels out and becomes irrelevant.  Earlier today i was working with area values and Pi (3.14) and got all goofed up. Im exhausted just thinking about this stuff.

    ---mac---

    mac
    • Member since
      September, 2006
    • From: New Jersey
    Posted by mac on Monday, August 22, 2016 4:05 PM
    mac
    • Member since
      September, 2006
    • From: New Jersey
    Posted by mac on Monday, August 22, 2016 4:24 PM

    • Member since
      July, 2005
    Posted by 1FASTXC on Thursday, August 25, 2016 9:22 AM

    Mac, not looking for an argument, but your formula is not correct. It doesn't figure out actual area, it just figures out the outside circumference of each diameter. to get area you need PI(3.1416)Xradius squared. That will give you area in either cubic centimeters or inches depending on what you use for units.

    "Slightly stock"
    • Member since
      July, 2005
    Posted by 1FASTXC on Thursday, August 25, 2016 9:34 AM
    Machz975, to answer your question I will use mac's example. If the bore is 3.5 and the inside of the squish band is 2.75 you would take 1.75(half the bore diameter) squared X PI(3.1416). Then 1.375 (half the squish inside diameter) squared X PI. Then subtract the second number from the first to give actual squish area and then divide second number by first number. See here. 1.75 X 1.75= 3.0625 X 3.1416= 9.6211. (your bore area in cubic inches) 1.375X1.375= 1.89 X 3.1416 = 5.939.(your dome area in cubic inches) 9.6211-5.939 = 3.6821 (Your squish band area in cubic inches) 3.6821/9.6211 = .3827 or 38% squish area compared to bore. I know it may be confusing, but it is accurate and I've used it many times over the years designing heads. Hope it helps.
    "Slightly stock"
    mac
    • Member since
      September, 2006
    • From: New Jersey
    Posted by mac on Thursday, August 25, 2016 12:41 PM

    No offence taken.    Yeah i missed the radius part of the equation.

    http://i176.photobucket.com/albums/w198/mac7777/squish/squish%20percentage4_zpswmgcyodt.jpg

     

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