three angles, two are 90 degrees, the third is an infinite fraction of 360
Oh, I see. It's an infinitely long/tall isocoles (sp?) triangle. Or infinitely narrow, which I think is the case you're using. I see how you're doing this.
x^2+y^2+z^2<r^2
it has to be a volume, and thus it must be an inequality
There's no volume there. Both sides are in units of area.
my point with these inequalitines is just than if we can find some two dimensional equivilant, then it may help up find a non calculus way to get the area of a circle.
This doesn't make sense to me.
this is however, probably an impossible task, so lets instead try to find a proof that no such means can exist.
Neat!
Originally posted by AThousandYoungThere's no volume there. Both sides are in units of area.
[b]three angles, two are 90 degrees, the third is an infinite fraction of 360
Oh, I see. It's an infinitely long/tall isocoles (sp?) triangle. Or infinitely narrow, which I think is the case you're using. I see how you're doing this.
x^2+y^2+z^2<r^2
it has to be a volume, and thus it must be an inequality
There's no volume ...[text shortened]... impossible task, so lets instead try to find a proof that no such means can exist.[/b]
Neat![/b]
Yeah, you really couldn't read off a measure of volume from the equation, but it does give an equation for a sphere (solid) of radius r.
my point with these inequalitines is just than if we can find some two dimensional equivilant, then it may help up find a non calculus way to get the area of a circle.
Would this kinda be like using a using a line integral to find area, only backwards? It's been a while since I did that stuff.
Originally posted by AThousandYoungYeah technically you're right. Although the difference is really negligible here. Notice the set {(x,y,z) : x^2 + y^2 + z^2 < r^2} is bounded but open. That infintesimally thin surface where this relation holds with equality has measure 0. So < and <= are basically describing the same solid.
[b]Yeah, you really couldn't read off a measure of volume from the equation, but it does give an equation for a sphere (solid) of radius r.
Oh, I see. It's a sphere smaller than one of radius r, though.[/b]
Originally posted by telerionbut one could say similarly of any area
Think of centering the circle on a Cartesian plane. Basically, in the limit as I -> infinite, you are adding up the "area" of an infinite number of rays extending from the origin to a point on the circle, that is from (0,0) to some (x,y) such that x^2 + y^2 = r^2.
OK this is horribly ad hoc, obviously prone to measurement error, and all around a silly (especially the end); but maybe it can get the juices flowing.
What if you took a rectangular solid with a known volume and filled it with water? Then you poured the water into a cylinder with known radius and circumfrence. Now posit that the volume of the cylinder is the area of the base (the circle) multiplied by the height of the cylinder (Not too much of a leap since this is the equation for the volume of the retangular solid).
Now measure the height of the water (best if done with glass cylinder).
With the volume, height, and base dimensions known, you can back out the area from assumed relationship, A=V/h.
So there you go. Anytime you have a circle, cut it out or whatever, make a cylinder with the circle as your base, pour a known volume of water into the cylinder, measure the height, and bam! Area of the circle.
Here is the crappy part. Now you have to figure out the relationship between the radius and the area.
No problem, Archimedes. Take repeated samples as described above. Then posit a nonlinear relationship A=Br^x. Take the log of both sides, and you get log(A)=x*(log(B+log(r))). Regress A on r and BAM! the equation of the area of a circle. Don't forget to convert back from logs.
🙂 Ok math nerds, I'm ready for my spanking now.
Originally posted by telerionNice!
So there you go. Anytime you have a circle, cut it out or whatever, make a cylinder with the circle as your base, pour a known volume of water into the cylinder, measure the height, and bam! Area of the circle.
Here is the crappy part. Now you have to figure out the relationship between the radius and the area.
No problem, Archimedes. Take r ...[text shortened]... on r and BAM! the equation of the area of a circle. Don't forget to convert back from logs.
Maybe you could weigh it, as it could be more accurate than measuring the height.
Regressing to get an equation? Doesn't make sense to me... How could you know if the relationship was correct to begin with?
Originally posted by PalynkaWeighing it is a good idea.
Nice!
Maybe you could weigh it, as it could be more accurate than measuring the height.
Regressing to get an equation? Doesn't make sense to me... How could you know if the relationship was correct to begin with?
It should be empirically clear that widening the circle increases the area. The real sketchy part is choosing the functional form A=Br^x. Plus regression is based on calculus so . . .
Still I was thinking. They should have been able to get an approximation of pi after a while just by noticing that dividing the circumfrence by the diameter always gave the same number (well with measurement error).
Now if any ancient with some time on his/her hands happened to take all these area samples and divide them by pi, the quadratic relation between radius and area should stick out like a sore thumb.
Would it be reasonable to make this intuitive leap? Sure. Since I'm assuming the ancients have noticed the relationship between circumference and radius, and that somehow these should relate to a circle's area (big radius <=> big circumfernce <=> big area), I think dividing by pi would come about pretty quickly.
I agree, the bigger proportional increases in area should immediately imply testing for a quadratic relation before anything else.
Anyway, I think the perimeter/diameter relation has to be the key.
If you have a line and want to convert it to a rectangle you multiply by the length of the other side. It makes sense, as you imagine infinite lines next to each other until the length of the other size is complete.
The problem with the circle is the need to overlap this lines in order to "fill" the gaps near the perimeter, so Pi could include a measure of this overlapping.
If there was no overlapping we could think of something like radius*perimeter=area. If we define perimeter as diameter*Pi then
Area = 2*r^2*Pi
That means a non-overlapping-circle would be twice the area of a real one.
We can now rule out that Pi includes this overlapping and in reality has nothing to do with the area, but merely with the relation of perimeter and diameter. Area = radius*perimeter/2
What was the point again? 🙂