[Guest]

A point "A" is marked on the circumference of a wheel of radius "r".

The wheel is placed such that it touches the ground (at origin) on point A. The wheel is moved (rolled) horizontally and the path of point A is traced (green curve in the picture below).

What is the length of this path (green curve) between 2 points where point "A" touches the ground?

PS: In case you are thinking too much about the picture, the green curve I drew is not to scale. Its just an illustration to clarify the problem.

[Guest]

it should be 8*r for the cycloid.

[Guest]

I guess ogden_tbsa is correct.

The longer answer:

It seems to me that point A will make half ofan ellipse, the horizontal distance traveld should be the perimiter ofthe circle, i.e. 2*pi*r and the vertical distance is the diameter ofthe circle, i.e. 2*r.

Therefore, the small radius of the ellipsewould be 2*r and the big one is pi*r, then if we use the simplifiedformula of determining the perimiter of an ellipse, the distancetraveled by point A would be pi*(2*r + pi*r)/2 = pi*r*(pi+2)/2 ~= 8*r

[Guest]

2(pi)r = path as it is the circ of the circle and also the diameter of the the new curve

(pi)r would be the radius of the new curve

2(pi)(pi)r would be the circ of the new circle

half of that would be (pi)(pi)r or the length of the new path, so

answer = r * (pi)^2

But it has been a long time and geometry wasn't my strength.

[Guest]

I think maybe.

Wouldn't the leangth of the line to be measured be equal to the circumfrence of the wheel if point "A" made a complete rotation?

[Pickett]

I think maybe.

Wouldn't the leangth of the line to be measured be equal to the circumfrence of the wheel if point "A" made a complete rotation?

Not quite...the distance along the X-Axis would be the circumference of the wheel...because that's how far point "A" travelled along the X-Axis...the question is asking the distance in the x-y plane, which is not quite the same thing...

[Guest]

Length is given by 8r. I'm not sure if it is exactly 8 or not because I did the calculation numerically, but I would guess it is.

The calculation I used was: r * 2^0.5 * int(0~2pi){root(1 - cos(x)) dx}

[Guest]

Galileo found that the length of the arc was 8 times the radius.

Fermat found that the area under curve was 3 * pi * r^2.

Both mathematicians predated the advent of integral calculus.