# Math Question: What is the max distance of x?

What is the maximum distance of x in order to limit the slope change to less than .15 degrees?
###### Who is Participating?

Commented:
If the curve is a circle, and if the angles are x = ±0.075 deg,
then the radius of the circle is  R = 1/sin(x) = 763.9439
and the height at the center is  R(1 - cos(x)) = 0.0006545
1

Commented:
It depends.
What is the curve?  A circle or a parabola or something else?
Are you launching a rocket or a car?
Are you shooting a bullet or a canon ball?  How fast is it going at the start?
And did you really mean 0.15 degrees?  That is such a small angle that it is difficult to measure it by hand.

The slope is slightly positive to the left at -1.  It is slightly negative to the right at +1.
A change of 0.15 degrees could mean that it starts at +0.075 degrees and changes to -0.075 degrees.  Is this what you want?
1

Is this the railroad track question ?

You can readily approximate by assuming you have two back to back triangles and work it out that way.

For a mile long railroad track with 1 foot inserted, x is about 44 feet
0

Developer AnalystCommented:
.0026
0

Commented:
0.15 degrees is pi/1200 radians. Since the slope of a tangent to any curve is dy/dx = tan(theta) then theta must be less than pi/1200. The value of x needed to be found is obtained by setting y to zero. Therefore we need two conditions y=0 and dy/dx to be less than tan(pi/1200) - at least in the range [-1,+1]. So what we now need to know is what y is as a function of the abscissa.

Given the curve is a circle. First the circle must cut the abscissa at -1 and +1 which means it is symmetric in x so that the equation is x² + (y+a)² = r² where r is the radius and a the y offset to the center. Since the center is below the x axis a must be negative. We immediately obtain the result (y+a)²=r²-1. Similarly we obtain dy/dx = -x/sqrt(r²-1) (by substituting (y+a) after differentiation. This must still be less than tan(pi/1200) but we still have TWO unknowns r and a.
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.