How to find the normal at a point on a circle

[SallyGreen]

does anyoneknow how to find the normal at a point on a circle, and how to find the x, y coordinate at this point, hope anyone could help

 

[Renge Ishyo]

The normal is perpendicular to the surface of the circle at any point (since the circle is a 2D surface in and of itself and lines drawn normal are perpendicular to the surface). The line sticks out of the page towards your eye. The midpoint of the circle can be found by drawing the diameter of the circle (the midpoint is at the center). Usually in physics (as in rotational kinematics and dynamics), the normal is drawn as a line perpendicular to the midpoint of this circle which is given the coordinate (0,0) as a "flat" 2D coordinate system is placed over the circle. From here, you can specify the location of any other line normal to the surface of the circle by giving the coordinates as usual.

 

[SallyGreen]

Thanks Renge appreciate your help, do we need to calculate the x and y component of the normal, so for a circle would be -x0 (the x coordinate of centre of the circle)and -sgrt(1-x^2)respectively, but why the minus sign?

 

[Defennder]

You need to specify at which point you want to find the normal (line or vector), as well as whether you want to find the equation of the normal line at that point or the normal vector. They are different. dy/dx gives you the gradient of the tangent line at a given point on the circle. And we know that m₁m₂ = -1 if m1 and m2 are the gradients of 2 lines perpendicular to each other on the plane. So see how to find the normal line?

For the normal vector, one approach would be to first find the normal line, then express it in a vector equation, then extract the vector component of the equation for the normal. A quicker way would be to evaluate grad(f) where f is the equation of the circle for the normal vector.

 

[SallyGreen]

does anyone guys advice me how to find the curvature at the corner of a rectangular,, cos I need to find it at any point of such geometry, and the curvature for flat side is just zero, but still sruggling with the corner....

anyone could help...

 

[Defennder]

Is the curvature defined for the "sharp" edges of graphs?

 

[tiny-tim]

corners

does anyone guys advice me how to find the curvature at the corner of a rectangular,, cos I need to find it at any point of such geometry, and the curvature for flat side is just zero, but still sruggling with the corner....

anyone could help...

Hi Sally!

Corners don't have curvature (or you could say they have infinite curvature … the opposite of zero curvature).

What exactly is the question you were set?

 

[SallyGreen]

Is the curvature defined for the "sharp" edges of graphs?

yeah the curvature for the "sharp" edges

 

[HallsofIvy]

yeah the curvature for the "sharp" edges

Which, as you have been told, does not exist. Curvature is only defined for "smooth" curvers- that is at places where the derivative exists.

 

[SallyGreen]

Hi Sally!

Corners don't have curvature (or you could say they have infinite curvature … the opposite of zero curvature).

What exactly is the question you were set?

Hi there, Many thanks
As I need to consider the contribution of the curvature of the boundary of a triangle which is zero on the sides, but what about at the corners? I guss the corner does have a curvature which is infinite curvature, but why have u taken the opposite of the curvature?

 

[SallyGreen]

You need to specify at which point you want to find the normal (line or vector), as well as whether you want to find the equation of the normal line at that point or the normal vector. They are different. dy/dx gives you the gradient of the tangent line at a given point on the circle. And we know that m₁m₂ = -1 if m1 and m2 are the gradients of 2 lines perpendicular to each other on the plane. So see how to find the normal line?

For the normal vector, one approach would be to first find the normal line, then express it in a vector equation, then extract the vector component of the equation for the normal. A quicker way would be to evaluate grad(f) where f is the equation of the circle for the normal vector.

Hi there

if the line equation is 5x+6y=8, we can extract the vector as (5,6) is that correct?

 

[tiny-tim]

I guss the corner does have a curvature which is infinite curvature, but why have u taken the opposite of the curvature?

Hi Sally!

I was thinking of a straight line as having (obviously) zero curvature, and a corner is as far away from a straight line as you can get … hence the "opposite" of straight … and the opposite of zero is infinity.

Also, the curvature is defined to be the inverse of the radius of curvature (eg, a straight line has infinite radius of curvature, and therefore zero curvature), and a corner has zero radius of curvature.

As I need to consider the contribution of the curvature of the boundary of a triangle which is zero on the sides, but what about at the corners?

But why?