is it possible to work out a radius with just a small piece of a curve?
969870
It's difficult to know your options without knowing how big the building is and what shape it is. If the walls are at right angles then the length of the curved corner will be a quarter or the circumference from which you can find the radius. The radius would be twice the length divided by pi.\nIf not then, aside from taking a mould and working with tangents, I can't think of an easy way.\nThe answer above will work very well if you have access to the face of the curve.
945543
A short element of the curve is dL=sqrt(dr^2+(r*dt)^2), r*dt elementary length of the arc with radius r and polar angle t (your theta), dr elementary increase of r for dt.\nThus dL=sqrt(r’^2+r^2)*dt.\nIf  r=a*(1-cost), then dL=a*sqrt((sint)^2+(1-cost)^2)=a*sqrt(2)*sqrt(1-cost)*dt. \nFortunately 1-cost=2*(sin(t/2))^2 or dL=2a*sin(t/2)*dt, L=-4a*cos(t/2) for t=0 until 2pi, thus L=-4*a*(-1-1)=8*a – Oh! check it poor kid!\nIf r=(sin(t/2))^2=0.5*(1-cost), then a=0.5 for t=0 until pi. It must be done likewise. Thus L=-4*0.5*(0-1)=2
424740
Unfortunately there is not. There is however an algorithm that will allow you to work out square roots by hand.
312528
The radius of a circle is the distance from the center of the circle to the edge of the circle.
141519
It is the length divided by the radius
950608
get some graph paper. plug in numbers for x and calculate its equivalent y value. do that 5 times, and you should have your curved line. If not, use higher or lower numbers than you used the first time ( I like starting with 0) Then  draw a line perpendicular to the most bent part (aka axis of symmetry, aka cut the line in two so both sides of the new line look like mirror images), find 2 points on that cutting line, and there you go.
1031485
No.\n\nA chord can be any length that is less than (or equal to) the diameter of the circle.\n\nYou don't have enough information in your question.
691057
Imagine cutting the tube down its length and opening it up into a long, narrow rectangle.  The outside area would be the length times the width.  The width of the rectangle is really the circumference of the circular cross section of the tube.  Thus, since you know the diameter, you have this:\n\nsurface area = length * pi * diameter\n\nwhere pi is approximately 3.14159265358979.\n\nNote, if the tube is not flexible, you have to be careful measuring the length of the bent tube.  it is the length of an imaginary line through the center of the tube, but as the other answerer pointed out, you can get this by taking the average of the "inside the curve" length and the "outside the curve" length.  Of course, if the tube is flexible, just straighten it out and measure it.\n\n\nNote--straightening, cutting and opening up, etc., will change the area slightly as it stretches part of the tube and compresses another part--this is a very small effect (unless you tie the tube into a tight knot or something) and probably can't easily be measured well anyway--should be too small to matter in most applications.
510334
x = a*sin(ang)\ny = b*cos(ang)\n\nang goes from 0 to 360 degress, a and b determine how oval it is (if a=b it is a circle)
450833
join the 2 pointz frm da arc vit da radius.measure out da value of da angle forned here.then substitute da values 4 this formula:  2*pie*radius/da angle u measured\n\nhappy solvin!
838513
Sigh - first guy had it right, but didn't explain it all.  \nPi = 3.14- all you need for the problem\nRadius= half the diameter of your circle. You find the midpoint and measure from there to the side of the circle, in any direction. Most problems give you the radius. or at least show you the midpoint on the circle.\n\nSo, it is 3.14 x the amount of the radius x 2= the area of the circle.\n\nYou can do this in reverse as well, if you know the area but not the radius.
602663
No, a line cannot intersect a circle in at most one point, unless you're talking about tangent. A line can be tangent to a circle at only one point.
316497
If a curve in n dimensions is given parametrically as:\n\n( f1(s), f2(s), ..., fn(s) ) for a <= s <= b\n\nthen you can approximate the arc length by dividing the domain [a,b] into many small pieces a = s0, s1, s2, ..., sk = b and summing the lengths of the chords (secants? straight line segments. whatever) between s_{i} and s_{i+1}.\n\nThe length of the curve is defined as the least upper bound of all such sums (if that set has an upper bound).\n\nUnder suitable hypotheses you can compute this as the definite integral of ``ds'' as s varies from a to b, using the Euclidean rule\n\n(ds)^2 = (df1(s)/ds)^2 + ... + (dfn(s)/ds)^2\n\ni.e., your length is the definite integral from a to b of the square root of the sum of the squares of the derivative of each coordinate function.\n\nFor example for a circle of radius one at the origin parameterized as ( cos(t), sin(t) ) as t varies from a = 0 to b = 2 pi you get ... 2 pi ... which is what you would expect.  :-)  :-)\n\nOr for the arc length of ( (s^3)/3 - s, s^2 ) from s = 5 to s = 8 the length is the integral from 5 to 8 of\n\nsqrt( (deriv (s^3)/3 - s w.r.t. s)^2 + (deriv s^2 w.r.t. s)^2 ) ds\n\n= sqrt( (s^2 - 1)^2 + (2s)^2 ) ds\n= sqrt( s^4 - 2s^2 + 1 + 4s^2 ) ds\n= sqrt( s^4 + 2s^2 + 1 ) ds\n= (s^2 + 1) ds\n\ni.e., integral from s=5 to s=8 of (s^2 + 1) ds.
820614
So am I answering a homework question for you? lol\n\nCircles have the minimal perimeter for the area contained.\nArea of a circle is pi x Radius x Radius. (pi R squared)\n\nNext best would be something close to a circle, so consider a hexagon or octagon, with the corners of the square intersecting the vertices of the outer figure.\n\nLet's try a circle first...\n\nDiameter of the circle would be the diagonal of the square, or five times the square root of two.  5 x 1.414 = 7.07.  So the Radius is 3.535 meters.\n\nPi x 3.535 x 3.535 = 3.14 x 12.5 = 39.34\n\nSo the circle solution is too big.\n\nNow work out on paper what an octagon or hexagon surrounding the square would look like.  One of those should be about 34 m square.  Remember the area of the square is already 25 m2, so you are looking for only 9 more m2.\n\nI played around for minute or so in paint, and came up with a parallelogram that worked too.  Finally, a smooth curving perimeter shape get you benefits similar to the "circle" solution, so an ellipse would be a good comparison solution.  Try that, and compare the perimeter result against the perimeter of the parallelogram, which has a perimeter of 24.23 meters.\n\nHave fun!\nOC
882048
Here's the answer you want:  With a compass set to the radius of the circle, start marking off radii around the circle, dividing it into 6 equal parts.  Connect any 3 alternate points and you'll have your triangle.\n\nNO MATH REQUIRED!
735957
Well. You can keep drawing tangents until one which you find is the steepest. From your info, i have the idea that the curve is in the shape of a bell. So perhaps you can try drawing tangents at the mid-range of the curve.\n\nHope this helps=)
840317
if it is a true arch the length would equal the diameter,\nso radius would be half the diameter (length)