Robust
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A mechanical engineer requires the turning of a circular disc giving an area of 16 units. What radius does he give the machinist?
The equation I would give is: sqrt area/sqrt pi = radius; thus, if an area 16 then 4/1.77245...= 2.2567...radius. r^2*pi = 16 areaFredGarvin said:What is an equation for area that involves the radius?
You have the area...You have a constant \pi...
Thanks for the confirmation, but I'm not out of the woods yet on this, for the problem as originally given (project engineer) did state the total surface area, the tolerance being incidental. The PE is one of those who likes to throw these kinds of questions at us, just to keep us on our toes I suspect.brewnog said:Looks good to me Robust.
Check that the given area is the cross sectional area, and not the total surface area of the disc. If this is so, you're spot on.
Don't be so quick to take abuse from contemporaries! Too many times I've lost face and backed down, only to find that I was right all along...
Robust said:But the implications of this one does present a serious conflict as regards the pi value, showing it to be irrelevant. the radius is given consistently regardless of the pi pi value employed (recognized pi values). Here is the given formula: sqrt area/sqrt pi = radius; r^2*pi = 16 area!
It's a hypothetical question. the thickness and other parameters are immaterial. Only the radius to the circular plane is required.brewnog said:If the PE is giving you the total surface area of the disc, this means you'll need to also know its thickness, - the total surface area comprising of two circular surfaces, and a 'strip' to go around the periphery.
Robust said:the problem as originally given did state the total surface area
Knitpicking is fine with me - probably the more the better!brewnog said:I'm not nitpicking here am I?
Not so simple considering the absurdity of the area to a closed continuum described by other than a whole number or ending decimal.pack_rat2 said:Never before has so much discussion devolved from the simple formula, A = pi*r^2...;)