🧵 What time will display one third?
Anonymous at Wed, 19 Jun 2024 05:48:53 UTC No. 16242241
A typical clock has 12 subdivisions, one for each hour. There are two arms, one for the hour and one for the minute.
In picrel, imagine that the minute hand is perfectly on the '6' subdivision. Then, the hour hand will be perfectly in between the '10' and '11' subdivisions. However, this doesn't produce a perfect angle which is a third of the whole, i.e. 360/3=120 degrees. Each arm is variable and affects the other's placement on the clock.
My question is this: at what time(s) will a clock's hour and minute hands form a perfect 120 degree angle?
I trust that the jannies will see that this question is too stupid to break rule 2. I'm just curious as to what level of math is required to solve this question. I genuinely can't figure it out.
Anonymous at Wed, 19 Jun 2024 06:31:25 UTC No. 16242271
4 o'clock? 8 o'clock? Did you even try to think about it?
Anonymous at Wed, 19 Jun 2024 06:53:32 UTC No. 16242277
>>16242241
The minute hand moves 360° for every 30° the hour hand moves (a twelfth). Start at exactly 12:00am to make things simple. Find when subtracting the angle of the hour hand from the angle of the minute hand equals 120°
120 = x - (x/12)
x = 1440/11
((1440/11)/360)*60 = 12:21:49am
The minute hand will form a second 120° angle within the hour when coming back towards the hour hand
((2*(1440/11))/360)*60 = 12:43:38am
Anonymous at Wed, 19 Jun 2024 11:25:02 UTC No. 16242483
>>16242277
those are only the first 2 solutions
minute hand does one rotation per hour so 12 rotations for single hour hand rotation
so there should be ~24 (23?) instances within 12 hours
how would you calculate the time or just angles of each instance?
Anonymous at Wed, 19 Jun 2024 11:34:38 UTC No. 16242494
>>16242241
hour clock rotates at rh = 1/12 per hour rate
minutes clock rotates at rm = 1 per hour rate
at midnight or midday, the clocks are aligned, ie the angles of the hours hand and the minutes hand are equal and both 0 (ah=am=0).
at any time t, ah(t) = rh * t, and am(t)=rm * t, where t is measured in hours.
you are looking for all cases where |ah(t)-am(t)| = 1/3 + k, where k is an integer. 2 cases come :
(1) ah(t)-am(t) = 1/3 + k OR (2) ah(t)-am(t) = -1/3-k = -1/3 + k (k's sign doesn't matter)
case 1 leads to :
rh * t - rm * t = 1/3 + k -> t * (rh-rm) = 1/3 + k -> t = (3k+1)/(3(rh-rm) )= -4*(3k+1)/11. Just plug values of k. It repeats every 11 consecutive integers.
case 2 leads to :
rm * t - rh * t = -1/3 + k -> t * (rm-rh) = -1/3 + k -> t = (3k-1)/(3(rm-rh)) = 4*(3k-1)/11. Same, just plug values of k. It repeats every 11 consecutive integers.
pic related show an example of values for case 1
Anonymous at Wed, 19 Jun 2024 12:22:16 UTC No. 16242539
>>16242241
How are there still homework threads during the summer?
Anonymous at Wed, 19 Jun 2024 12:24:21 UTC No. 16242543
>>16242539
Nevermind, sorry
Anonymous at Wed, 19 Jun 2024 12:27:12 UTC No. 16242545
>>16242483
For subsequent hours add 30 each hour to account for the hour hand moving
120 = x - ((x/12)+30)
x = 1800/11
((1800/11)/360)*60 = 1:27:16am
But, add only 21m 49s to get the second 120° time in each hour
1:27:16am + 00:21:49 = 1:49:05am, which conforms with >>16242494
– – – – –
120 = x - ((x/12)+60)
x = 2160/11
((2160/11)/360)*60 = 2:32:43am
+ 21m 49s = 2:54:32, again conforming with >>16242494
Anonymous at Wed, 19 Jun 2024 14:20:54 UTC No. 16242602
Hour hand goes 5 'min' every hour i.e 60minutes. One third rotation is 20 min.
x*5/60+20min=x
x(1-5/60)=20min
x(55/60)=20min
x=12*20/11 min
x=240/11 min
x=21 9/11 min