Surface area of a cylindrical rigid rod
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    Default Surface area of a cylindrical rigid rod

    Basically , I have two rods of 1cm diameter and 5 cm height. One of the rods is fabricated in such a way that a helical thread is created throughout the rod with a helix angle of 60° . Other rod is fabricated in such a way that the surface is diamond grit knurled throughout.

    So this is done to increase the surface area compared to the smooth rod.

    My question is that I want to find the surface area of each of the rods .

    I hope someone could help me with this.

    Thank you so much.

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    For starters, the basic formula for the area of a cylindrical surface is just the circumference times the length:

    A = pi x d x l

    A = the area
    pi = 3.141 (a constant)
    d = the diameter
    l = the length

    The 60 degree thread case:

    Several assumptions must be made. First, that 60 degree thread has a sharp crest and a sharp valley. Second, the thread depth or linear pitch is small in relation to the diameter. Finally, the OD (outside diameter) at the crest of the thread remains the same as the original OD of the cylinder.

    A SHARP, 60 degree thread viewed in cross section forms a equilateral triangle: the pitch line from crest to crest and the two flanks of the thread form the triangle and all three of these sides are equal. So, each flank of the thread has the same width and length and therefore the same area as the original surface of the cylinder (which has been cut out). Therefore the area of the surface has been multiplied by a factor of 2, for the two thread flanks. And the modified formula is:

    A = 2 x pi x d x l

    In case you are wondering, the pitch of the thread does not enter into this calculation. If my first assumption is not true and there is a flat at the crest of some kind of fill in the thread's root, then the area will be somewhat less. The exact amount less will depend on the size and shape of those features.

    Another flaw in this calculation is the second assumption: that the linear pitch of the thread is small compared to the cylinder's diameter. The formula is 100% accurate if the OD is infinite. Any smaller OD will mean that the parts of the thread flanks that are closer to the root of the thread sill be somewhat smaller due to the decreasing length of the thread at that, lower point. This decrease in length is due to the smaller diameter there. If the assumption is true, this error will be small and in many cases can be neglected. But if it is not true, then a more complicated calculation would be called for.

    I had other things to do and had to interrupt my work on this at this point. I will post about the other case, with the diamond pyramids, later.

    PS: If you are simply trying to maximize the surface area, among the three cases you cited, this is it. If you want even more, just decrease the included angle of the thread: the smaller, the better. Mathematically, there is no limit to the process: in theory, when the included angle of the thread approaches zero, the area approaches infinity.

    PS #2: It is the included angle of the thread, not it's depth that counts. Changing the depth of the thread is equivalent to changing it's pitch and the area remains the same for a given included angle. Within the limitations of assumption #2 anyway.

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    This took a while. I hope there are no mistakes. Perhaps someone will be nice enough to check it.

    The diamond pattern case:


    This case is more complicated. The angles of these diamond/pyramid shaped protrusions will be dependent on all of the angles used in creating them. In the threaded case I compared the surface area of a length of the thread to the original surface area above that thread. My calculation here will be along the same lines: the area of the four sides of the pyramids compared to the base area of those same pyramids. I will be finding a ratio between these two and using that ratio to modify my formula for the simple cylinder.


    And again, several assumptions must be made. First, that diamonds have a sharp crest and a sharp valley. Second, the depth small in relation to the diameter. Finally, the OD (outside diameter) at the crest of the diamonds remains the same as the original OD of the cylinder. These are quite similar to the assumptions I made in the first case. And when you cut or roll a diamond pattern, there are two pitches involved, one for each pair of sides of the diamonds. In most cases these two pitches are equal and that means that the angles of the two pairs of sides are the same, just in opposite directions. But this is not a necessity. for simplicity, I will be assuming that they are equal and the two helical angles will be the same. The answers would be similar in any case.

    And, like a multi start thread, there are two linear distances along the thread axis that are involved: the pitch and the lead. The pitch is the distance from one crest to the next. This is true for any thread, single or multi start. The lead is the distance that the thread travels with one full turn. For a multiple start thread, the lead is a multiple of the pitch. Example: a three start thread with a pitch of 0.050" will have a lead of three times that or 0.150". It is important to keep this in mind because the angle Ha which I define next is the LEAD angle of the helix. A calculation using this angle and the circumference of the cylinder will give us the LEAD of the helix, not the distance across a single diamond/pyramid. That distance would be the pitch and it will be an even division of the lead. It is that pitch that we need to use in our calculations. This is not an assumption, it is a fact that must be kept in mind.

    As a practical matter, you can determine the number of starts for the diamond pattern by just counting the number of diamonds along a circumference of the cylinder. If the two helix angles are the same, this will be a whole number. But if knurling wheels are used to create the pattern, there is always the chance that the angles are a little different and the count of diamonds may involve a 1/2 or even a 1/3 fraction. If this is true, then it will introduce an additional error in the process, but for approximate calculations it can probably be neglected. There would be a danger in including the fraction in your calculations because it may apply in either direction, plus or minus. And it would be difficult to tell which at a fast glance.

    n = number of starts

    There are two angles involved. First when we look down vertically at the surface we see a four sided outline of the diamonds. This will be a four sided, parallelogram shape and it will have two angles at the different corners. The smaller of these two angles is the one I will work with. Half of this smaller angle is the helix angle and this is the angle we need to work with.

    Ha = the helix angle = 1/2 the smaller parallelogram angle.

    The second angle is seen in a cross sectional view of one of these pyramids. This cross sectional view should be taken in a plane that is perpendicular to one of the base sides of the parallelogram base and that also goes through the peak. To save division by 2 in the math that follows, I am defining this as 1/2 of the full, included angle between opposite sides of the pyramid.

    Va = one half the included angle between opposite faces of the pyramids.

    p = linear pitch of the helix that created the pyramids = the distance across one pyramid.


    As a practical matter, the linear pitch (p) can be measured by tracing one or more revolutions of the helix between the pyramids and dividing the distance along the cylinder's axis involved by the number of starts multiplied by the number of revolutions used, Once the pitch (p) is known, you can find the helix angle (Ha) from the circumference (c) or the diameter (d) and the arctangent of the ratio between the lead (n x p) and the diameter.

    Ha = arctan(l / c) = arctan((n x p) / (pi x d))

    And finally, the height of the pyramids:

    h = vertical (or normal) height of the pyramids.

    This would be a shop measurement and could be a bit tricky.

    OK, here goes the area calculations.

    Base area of a pyramid. This is representative of the plain cylinder with no modifications:

    Ab = p x PyramidLength / 2

    The PyramidLength = c / n = pi x d / n

    So

    Ab = p x pi x d / 2n

    That was fairly easy. Now for the area of the sides of the pyramids.

    As = Area of the four pyramid sides.

    A1 = Area of one pyramid side.

    As = 4 x A1

    The sides are triangles so the area is the one half the base times the height. The base is found from the helix angle, the diameter, and the number of starts.

    b = Base length

    b = (pi x d) / (n x cos(Ha))

    The height of the triangle is the height of the pyramids divided by the cosine of the included half angle (Va).

    th = Height of Side's triangles

    th = h / cos(Va)

    So the area of one side is:

    A1 = 0.5 x b x th

    A1 = 0.5 x ((pi x d) / (n x cos(Ha))) x (h / cos(Va))

    And the area of all four sides:

    As = 2 x ((pi x d) / (n x cos(Ha))) x (h / cos(Va))

    Finally, the ratio of the areas of the base of the pyramids to their four sides is:

    R = As / Ab

    And if you really want it all in one expression;

    R = (2 x ((pi x d) / (n x cos(Ha))) x (h / cos(Va))) / (p x pi x d / 2n)

    That will be more than 1 and less than the value of 2 that I got for the 60 degree threaded case. As I said above, the exact value will depend on those angles. I leave it up to you to take the measurements and find the actual ratio. I get dizzy just looking at all those brackets.

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    Holy Cow EPA, that was a lot of work. I hope the guy you're doing homework for appreciates it!

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    Zaqf, why do you want to know?
    I'm intrigued!
    fusker

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    Quote Originally Posted by fusker View Post
    Zaqf, why do you want to know?
    I'm intrigued!
    fusker
    Part of an assignment, I suspect...

    Dave H. (the other one)

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    Quote Originally Posted by Milland View Post
    Holy Cow EPA, that was a lot of work. I hope the guy you're doing homework for appreciates it!
    Disrespectful little sod:
    Surface area of a cylindrical rigid rod

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    Just guessing here but couldn't the surface area be determined empirically using displacement? Personally, if someone asked me to figure that out I would tell them to do it themselves. Benefit of getting old.

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    Sounds like the OP is trying to calculate how much carbide grit he needs to lay on the OD of a stablizer body used on drilling tools in the oilfield. If so, there's a lot of missing information that is needed for the calculations. Ken

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    Start by calculating the perimeter of the Koch snowflake and working backwards from there.....

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    Quote Originally Posted by MrStretch View Post
    Start by calculating the perimeter of the Koch snowflake and working backwards from there.....
    Be easier to start from the area of a Sierpinski gasket, surely?

    Dave H. (the other one)

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    Quote Originally Posted by MrStretch View Post
    Start by calculating the perimeter of the Koch snowflake and working backwards from there.....
    Quote Originally Posted by Hopefuldave View Post
    Be easier to start from the area of a Sierpinski gasket, surely?

    Dave H. (the other one)
    I am proud to award these qty (2) the coveted "JRIowa memorial Turboencabulator award"
    Last edited by digger doug; 01-29-2020 at 05:15 PM.

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    I read the Henry Ford descendants always got their engineering degree assignments done by professionals,and handed them in with the bill included,just so the lecturer could know it was quality work.

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    Displacement is a method that can be used to determine volume. Not area. Not surface area.

    Sorry bout dat!



    Quote Originally Posted by crossthread View Post
    Just guessing here but couldn't the surface area be determined empirically using displacement? Personally, if someone asked me to figure that out I would tell them to do it themselves. Benefit of getting old.

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    Quote Originally Posted by EPAIII View Post
    For starters, the basic formula for the area of a cylindrical surface is just the circumference times the length:

    A = pi x d x l

    A = the area
    pi = 3.141 (a constant)
    d = the diameter
    l = the length

    The 60 degree thread case:

    Several assumptions must be made. First, that 60 degree thread has a sharp crest and a sharp valley. Second, the thread depth or linear pitch is small in relation to the diameter. Finally, the OD (outside diameter) at the crest of the thread remains the same as the original OD of the cylinder.

    A SHARP, 60 degree thread viewed in cross section forms a equilateral triangle: the pitch line from crest to crest and the two flanks of the thread form the triangle and all three of these sides are equal. So, each flank of the thread has the same width and length and therefore the same area as the original surface of the cylinder (which has been cut out). Therefore the area of the surface has been multiplied by a factor of 2, for the two thread flanks. And the modified formula is:

    A = 2 x pi x d x l

    In case you are wondering, the pitch of the thread does not enter into this calculation. If my first assumption is not true and there is a flat at the crest of some kind of fill in the thread's root, then the area will be somewhat less. The exact amount less will depend on the size and shape of those features.

    Another flaw in this calculation is the second assumption: that the linear pitch of the thread is small compared to the cylinder's diameter. The formula is 100% accurate if the OD is infinite. Any smaller OD will mean that the parts of the thread flanks that are closer to the root of the thread sill be somewhat smaller due to the decreasing length of the thread at that, lower point. This decrease in length is due to the smaller diameter there. If the assumption is true, this error will be small and in many cases can be neglected. But if it is not true, then a more complicated calculation would be called for.

    I had other things to do and had to interrupt my work on this at this point. I will post about the other case, with the diamond pyramids, later.

    PS: If you are simply trying to maximize the surface area, among the three cases you cited, this is it. If you want even more, just decrease the included angle of the thread: the smaller, the better. Mathematically, there is no limit to the process: in theory, when the included angle of the thread approaches zero, the area approaches infinity.

    PS #2: It is the included angle of the thread, not it's depth that counts. Changing the depth of the thread is equivalent to changing it's pitch and the area remains the same for a given included angle. Within the limitations of assumption #2 anyway.


    While you're at it, EPAIII, can you tell me how old the bus driver is? Thanks.

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    I agree with Milland, impressive piece of calculation by EPA III there. In the other thread Digger Doug linked, also some nice CAD work. There were a couple of details that caught my eye in the OP, though, which may have been overlooked FWIW. The term "diamond GRIT knurled" was interesting, as well as the spec (one of very few) of the HELIX angle (not lead angle) of 60 degrees, which is more like a twist drill than a screw thread. If the actual calculations needed require a set of assumptions about the RMS average surface finish from a diamond-grit knurled surface, things may get a little more complicated....

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    Quote Originally Posted by Ron Hofer View Post
    While you're at it, EPAIII, can you tell me how old the bus driver is? Thanks.
    The long or the short bus?

    Dave H. (the other one)

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    Thanks for share.

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    The surface area of a modified cylinder is the profile length of the cylinder times its circumference.

    For example, if a perfectly round thread with a pitch of 0.1cm is cut out of the 5cm cylinder then the profile will be a series of 25 semicircles. Each semi-circle has a perimeter of 0.1 * 3.14 = 0.314cm. So, the total profile length is 0.314 * 25 = 7.85cm. If we multiply that by the circumference of the cylinder (3.14cm) then the total surface area is approximately 7.85 * 3.14 = 24.649 square centimeters.

    If your thread has some other shape, like Acme, then you would have to compute the length of the profile in an analogous way.

    For the knurl, you do the same thing: draw the profile of the knurl and estimate its length, then multiply by the circumference of the cylinder (3.14cm).


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