S2T2

Jim Randell 5:35 am on 18 August 2024 Permalink Reply
Tags: by: Howard Williams ( 43 )   

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  Jim Randell 5:36 am on 18 August 2024 Permalink | Reply

  By considering two right-angled triangles we can derive a linear equation for the depth of the centre of the arc below ground level (x), from which the width of the polytunnel (2w) can be determined.

  The triangles are indicated in the following diagram:

  

  This Python program (appropriately enough) uses the [[ Polynomial() ]] class from the enigma.py library to perform the calculations.

  It runs in 76ms. (Internal runtime is 139µs).

  Run: [ @replit ]

  from enigma import (Rational, Polynomial, sq, sqrt, printf)
  
  Q = Rational()
  
  # suppose the polytunnel has ground-level semi-width w
  # and the centre of the circle is a distance x below ground
  # we have:
  #
  #  r = x + 300
  #  r^2 = w^2 + x^2
  #  r^2 = (7w/11)^2 + (225 + x)^2
  
  fx = Polynomial('x')  # depth of origin
  fr = fx + 300  # radius of arc
  
  f1 = sq(fr) - sq(fx)  # = w^2
  f2 = sq(fr) - sq(fx + 225)  # = (7w/11)^2
  
  f = sq(Q(7, 11)) * f1 - f2  # equate values of w^2
  
  # solve f(x) = 0 for positive real x
  for x in f.roots(domain='F', include='0+'):
    # calculate width
    W = sqrt(f1(x)) * 2
    printf("polytunnel width = {W:g} cm [x = {x:g}]")
  

  Solution: The width of the polytunnel at ground level is 660 cm.

  The polynomial to determine the depth of the centre of the circular arc simplifies to:

  f(x) = 2x − 63

  From which we easily find the depth of the centre and the radius of the circle:

  x = 31.5
  r = 331.5

  And from these we calculate the semi-width of the polytunnel:

  w^2 = r^2 − x^2 = (r + x)(r − x)
  w^2 = 363 × 300 = 108900
  ⇒ w = 330

  So the total width of the polytunnel is 660 cm.

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  Frits 11:24 am on 18 August 2024 Permalink | Reply

  I made my narrowing down routine from scratch.
  Of course there are more efficient ways to do this (I don’t know them by heart).

  from sys import float_info
  eps = float_info.epsilon
  
  # r = x + 300    # the centre of the circle is a distance x below ground
  # 49.w^2 - 117600.x - 17_640_000 = 0
  # 49.w^2 -  72600.x - 19_057_500 = 0
  
  f1 = lambda x: 17_640_000 + 117_600 * x
  f2 = lambda x: 19_057_500 +  72_600 * x
  
  mode = ""
  x = 100         # choose an arbitrarily starting value
  delta = 5       # increment/decrement value for variable x
  
  # find a value for "x" where both f1(x) and f2(x) are (almost) the same
  while True:
    v1, v2 = f1(x), f2(x)
    # do we have an solution where both values are (almost) the same?
    if abs(v2 - v1) <= eps:
      w = (2_400 * x + 360_000)**.5
      print(f"answer:  ground-level width = {round(w, 2)} cm")
      break
    
    if v1 < v2:
      if mode == "more": 
        delta /= 2
      x += delta
      mode = "less"
    else:
      if mode == "less": 
        delta /= 2
      x -= delta
      mode = "more"     
  

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  • 

    Frits 11:34 am on 18 August 2024 Permalink | Reply

    The program was only made if you don’t want to analyse the problem till a solution is found.

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  GeoffR 6:23 pm on 18 August 2024 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  % X = half tunnel width, Y =  depth below ground, R = circle radius
  % All dustances in mm.
  
  % Assumed UB's for variables
  var 1..4500:R;  var 1..4000:X; var 1..1500:Y; 
  
  constraint R == 3000 + Y;  
  
  constraint X * X  + Y * Y == R * R;
   
  % (2250 + Y)^2 + (7/11 * X)^2 = R * R
  constraint 121 * R * R == 121 * ((2250 + Y) * (2250 + Y)) + (49 * X * X);
  
  solve satisfy;
  
  
  

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