When the Swiss didn't have so much experience yet with making clocks, a painful mistake was made with a church clock. The clock was officially put into use when it showed 6 o'clock. But soon it was noticed that the hour hand and minute hand had been interchanged and attached to the wrong axes.
The result was that the hour hand moved with a speed precisely 23 times higher than the minute hand. When the clock maker arrived, a remarkable thing happened: on the moment he inspected the clock, it showed exactly the right time again.
If the clock started at 6 o'clock in the correct position, then what was the first moment that it showed the correct time again?
I did this fairly quickly (for me), so hopefully I didn't make an error.
With my methodology, I get 6:28, just 28 minutes after 6.
initial conditions:
m = 0
h = 180
Vm = v degrees per minute
Vh = 23v degrees per minute
We only know the relative speeds, not the actual speed.
I expect multiple answers scaling the absolute speed so that the correct time could be at various future "legal positions" of the hands.
To be a legal position, the hour hand must be the same percent of the way between 2 adjacent numbers as the minute hand is around the entire circle.
So, m/360 = mod(h,30)/30
Define minute hand position in terms of hour hand, so there is only one solution for any value of h
legal_m(h) = 12*mod(h,30) where legal_m and h are in degrees
faulty_m(t) = (0 + v*t) mod 360
faulty_h(t) = (180 + 23*v*t) mod 360
From the output of the program below, it appears that every 14.4 "time periods" there is a valid time showing on the clock. But we want the time that is after the 6 o'clock position by the least amount.
So rather than choose the first solution which is 14.4 time periods later, I'll choose 172.8 time periods after 6 o'clock, at which point the clock reads 6:28:00
So I need the speed of the minute hand to be 172.8 time periods in 28 minutes or 6.1714285 minutes on the clock for every minute of real time, or 37.0285714 degrees per minute.
If there is a known absolute speed of the two hands, then there will be a different answer.
def legal_m(h):
""" where legal_m and h are in degrees """
return 12*(h%30)
def faulty_m(t):
""" where faulty_m is in degrees, t is in minutes; assume v=1 """
v=1
return (v*t)%360
def faulty_h(t):
""" where faulty_h is in degrees, t is in minutes; assume v=1 """
v=1
return (180 + 23*v*t)%360
dt = 100 # there are dt units in one minute
epsilon = .001
for time in range(200 * dt):
t = time/dt
fm = faulty_m(t)
fh = faulty_h(t)
if abs( fm - legal_m(fh)) < epsilon:
print(t, fh, fm, legal_m(fh))
print('the time reads {}:{}:{}'.format(int(fh/30) , int(fm/6), round(10*fm%6) ))
print()
OUTPUT:
0.0 180.0 0.0 0.0
the time reads 6:0:0
14.4 151.2 14.4 14.399999999999864
the time reads 5:2:0
28.8 122.39999999999998 28.8 28.799999999999727
the time reads 4:4:0
43.2 93.59999999999991 43.2 43.19999999999891
the time reads 3:7:0
57.6 64.79999999999995 57.6 57.599999999999454
the time reads 2:9:0
72.0 36.0 72.0 72.0
the time reads 1:12:0
86.4 7.199999999999818 86.4 86.39999999999782
the time reads 0:14:0
100.8 338.4000000000001 100.8 100.80000000000109
the time reads 11:16:0
115.2 309.5999999999999 115.2 115.19999999999891
the time reads 10:19:0
129.6 280.7999999999997 129.6 129.59999999999673
the time reads 9:21:0
144.0 252.0 144.0 144.0
the time reads 8:24:0
158.4 223.20000000000027 158.4 158.40000000000327
the time reads 7:26:0
172.8 194.39999999999964 172.8 172.79999999999563
the time reads 6:28:0
187.2 165.59999999999945 187.2 187.19999999999345
the time reads 5:31:0
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Posted by Larry
on 2023-04-28 12:05:04 |
Solution ?
