How many points are need to construct a smooth circular arc?

Question

Hi all.  I need to construct circular arc by  array using Arcpy. Apparently the larger the circle, the more number of points in ArcMap. How many points are need to keep same smoothness with ArcMap platform, what's the relationship between diameter R and Number of points to get it?

DanPatterson_Retired · Answer

follow up   an arc between 0 and 5 degrees, center at 5,5, 50 points... using the np.linspace option

>>> ang=np.linspace(0, 5, num=50, endpoint=True, retstep=False)
>>> ang= np.deg2rad(ang)
>>> Xs = 1*np.cos(ang)
>>> Ys = 1*np.sin(ang)
>>> pnts = np.array(list(zip(Xs,Ys)))
>>> pnts = np.round(pnts + [5,5], 3)
>>> pnts
array([[ 6.000,  5.000],
       [ 6.000,  5.002],
       ...
       [ 5.997,  5.080],
       [ 5.997,  5.082],
       [ 5.996,  5.084],
       [ 5.996,  5.085],
       [ 5.996,  5.087]])

much finer resolution, over a 5 degree range.  Using this option you are pretty well limitless.

DanPatterson_Retired · Answer

up for some numpy and python?

# -*- coding: UTF-8 -*-
"""
:Script:   circle_make.py
:Author:   Dan.Patterson@carleton.ca
:Modified: 2016-08-03
:Purpose: make a circle
"""
#---- imports, formats, constants ----
import sys
import numpy as np
from textwrap import dedent

ft={'bool':lambda x: repr(x.astype('int32')),
    'float': '{: 0.3f}'.format}
np.set_printoptions(edgeitems=10, linewidth=80, precision=2,
                    suppress=True, threshold=100, 
                    formatter=ft)
script = sys.argv[0]

#---- functions ----
def _demo(radius=100, theta=1, xc=0.0, yc=0.0):
    """   
    :Requires
    :--------
    :  radius - in projected units
    :  angle - integer
    :Returns
    :-------
    :  list of coordinates rounded to 3 decimal places
    :Notes:
    :------
    : If you need more than 360 points use, for example 720 points
    : np.linspace(start, stop, num=50, endpoint=True, retstep=False)
    : np.linspace(-180, 180, num=720, endpoint=True, retstep=False)
    """
    angle = [np.deg2rad(i) for i in range(-180,180,theta)]
    Xs = radius*np.cos(angle)         # X values
    Ys = radius*np.sin(angle)         # Y values
    pnts = np.array(list(zip(Xs,Ys)))
    pnts = np.round(pnts + [xc,yc], 3)
    p_lst = pnts.tolist()
    return p_lst

#----------------------
if __name__=="__main__":
    """   """
    #print("Script... {}".format(script))
    p_lst = _demo(radius=1, theta=30, xc=5, yc=5)
    print(p_lst)

Now if I return the pnts array instead of the list (p_lst) then the output for the demo with a center (5,5) radius 1 and angle of 30

array([[ 4.000,  5.000],

[ 4.134,  4.500],

[ 4.500,  4.134],

[ 5.000,  4.000],

[ 5.500,  4.134],

[ 5.866,  4.500],

[ 6.000,  5.000],

[ 5.866,  5.500],

[ 5.500,  5.866],

[ 5.000,  6.000],

[ 4.500,  5.866],

[ 4.134,  5.500]]))

in list form

[[4.0, 5.0], [4.134, 4.5], [4.5, 4.134], [5.0, 4.0], [5.5, 4.134], [5.866, 4.5], [6.0, 5.0], [5.866, 5.5], [5.5, 5.866], [5.0, 6.0], [4.5, 5.866], [4.134, 5.5]]

If you need more than 360 points, see the notes in the header section and generate the angles using np.linspace

TieshengWu · Answer

Thank you Dan. below is code:

def circle_array(x,y,c1,c2,r):     #x,y for center, c1,c2 for start and end angle, r for diameter

ar=[]

deg2rad=math.pi/180.0

n=360.0

for i in range(0,n+1):

ang = (c1+(c2-c1)*i/n) * deg2rad

arx=x + r * math.cos(ang)

ary=y + r * math.sin(ang)

ar.append([arx,ary,0])

return ar

I just wonder how to decide the n to keep the same smoothness with arcmap platform