Sunday Times Teaser 2954 – Lovely Meter, RITA Made!

by Stephen Hogg

Published May 5 2019 (link)

Revisiting the old curiosity shop, I bought an unusual moving-iron ammeter (made by Rho Iota Tau Associates). On the non-linear scale “9” was the last possible “whole amperes” scale graduation marked before the “full scale deflection” end stop (which was a half turn of the pointer from zero).

The booklet gave the pointer swing from zero (in degrees) equal to the current (in amperes) raised to a fixed single-figure positive whole number power, then divided by a positive whole number constant. Curiously, the angle between “exact” pointer swings, calculated using this formula, for two single-figure “whole amperes” values was exactly a right angle.

What, in amperes, were these two currents (lower first) and to what power were they raised?

4 Comments Leave one →

Brian Gladman permalink


from itertools import combinations

# the power involved
for p in range(1, 10):
  # look for two single digit values giving a 90 degree difference
  # in their angles, i.e. (b^p - a^p) == 90.C for integer C
  for a, b in combinations(range(10), 2):
    C, r = divmod(b ** p - a ** p, 90)
    # and with the 9 and 10 amp angles being either side of 180 degrees
    if not r and 9 ** p / C < 180 < 10 ** p / C:
      print(f"{a} and {b} amps (degrees = amp^{p} / {C}).")

from itertools import combinations

# the power involved

for p in range(1, 10):

# look for two single digit values giving a 90 degree difference

# in their angles, i.e. (b^p - a^p) == 90.C for integer C

for a, b in combinations(range(10), 2):

C, r = divmod(b ** p - a ** p, 90)

# and with the 9 and 10 amp angles being either side of 180 degrees

if not r and 9 ** p / C < 180 < 10 ** p / C:

print(f"{a} and {b} amps (degrees = amp^{p} / {C}).")

Erling Torkildsen permalink


from itertools import permutations
for e in range(2, 10):
  for d in range(1, 1000000):
    # 9 amps gives an angle less than 180 degrees 
    if 9 ** e / d < 180:
      a = []
      # compute possible angle differences for integer currents
      for n in range(1, 10):
        a.append((n ** e) / d)      
      ax = permutations(a, 2)
      # pick right angles
      for a1, a2 in ax:
        if 90 == int(a2 - a1) == a2 - a1:
          I1, I2 = int((d * a1) ** (1 / e)), int((d * a2) ** (1 / e))
          print(f"{I2}^{e}/{d} - {I1}^{e}/{d} = {a2} deg - {a1} deg = {int(a2-a1)} deg.")

from itertools import permutations

for e in range(2, 10):

for d in range(1, 1000000):

# 9 amps gives an angle less than 180 degrees

if 9 ** e / d < 180:

a = []

# compute possible angle differences for integer currents

for n in range(1, 10):

a.append((n ** e) / d)

ax = permutations(a, 2)

# pick right angles

for a1, a2 in ax:

if 90 == int(a2 - a1) == a2 - a1:

I1, I2 = int((d * a1) ** (1 / e)), int((d * a2) ** (1 / e))

print(f"{I2}^{e}/{d} - {I1}^{e}/{d} = {a2} deg - {a1} deg = {int(a2-a1)} deg.")

Erling Torkildsen permalink

A critical comment on my Python-program above.
The angles in question are given by the three variables: the current (I), and a ‘calibrating’ exponent (p) and divisor (n), all integers. While I and p are limited to one digit, no limit is given for n.
The only sensible thing to do is therefor to pick I and p as the free variables and test if a corresponding n exists.
My programme (above) fails to do that. It was written spontaneously and gave a solution directly, that flattered the author and weakened his critical sense..

Reply

GeoffR permalink


from itertools import combinations

for a, b in combinations(range(2, 10), 2):
  for p in range(2, 10):
    # calculate the ammeter number constant(q)
    q, r = divmod (b **p - a ** p, 90)
    if r != 0: 
      continue
    # check that 9 is on the scale and 10 is not
    if 9 ** p / q  < 180 < 10 ** p / q:
      print("Current values are {} and {} amps ".format(a, b))
      print("Raised power = {}, constant = {}".format(p, q))

from itertools import combinations

for a, b in combinations(range(2, 10), 2):

for p in range(2, 10):

# calculate the ammeter number constant(q)

q, r = divmod (b **p - a ** p, 90)

if r != 0:

continue

# check that 9 is on the scale and 10 is not

if 9 ** p / q < 180 < 10 ** p / q:

print("Current values are {} and {} amps ".format(a, b))

print("Raised power = {}, constant = {}".format(p, q))

Sunday Times Teaser 2954 – Lovely Meter, RITA Made!

by Stephen Hogg

Published May 5 2019 (link)

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