At what time (to the second) are the hour and minutes hands aligned between 7:00 and 8:00?
So... 2 positions, opposite and superimposed.
6 comments
Let A be the angle of the hour hand and B the angle of the minute hand, in degrees.
Between the hour of 7 and 8,
A = 7 * 30 + (B / 12)
If they're the same, we can substitute A for B:
A = 7 * 30 + (A / 12)
A * 11/12 = 7 * 30
A = 229.09090909... degrees
So the time comes out to 7:38:10.909090...
If the clock "ticks", the closest time is 7:38:11.
Feel free to do the same method for the "opposite hands" case, I'm on mobile.
How do you start setup of a problem like this? I tried imagining both sides of an equation with a variable for "degrees per second", multiplied by current position.
Setting them equal kind of wrote me into a corner.
So this is a problem with two unknowns (the two angles of the hands). You need two different equations to solve it--in this case, one of the equations is simply A == B. The other is calculating A and B as an angle as a factor of time, and it turns out that B / 360 is the fraction of the hour that has gone by. So multiplying that fraction by the smaller 30 degrees that the hour hand travels in that same time gives us the number of degrees past the 7, then add the other 7 hours it's already traveled, and you have that relationship equation I posted.
Indeed! For opposite, B=A-180.
Gives A * 11/12=7 * 30-15
A=212.727272etc degrees.
So 7:05:27.2727...
Roughly 32min44s between the two occurrences, and that's true for every subsequent occurrence. This value can also be derived by considering there are 21 alignments between noon and midnight (not counting these), so dividing 720 minutes by 22 intervals
Let A be the angle of the hour hand and B the angle of the minute hand, in degrees.
Between the hour of 7 and 8,
A = 7 * 30 + (B / 12)
If they're the same, we can substitute A for B:
A = 7 * 30 + (A / 12)
A * 11/12 = 7 * 30
A = 229.09090909... degrees
So the time comes out to 7:38:10.909090...
If the clock "ticks", the closest time is 7:38:11.
Feel free to do the same method for the "opposite hands" case, I'm on mobile.
How do you start setup of a problem like this? I tried imagining both sides of an equation with a variable for "degrees per second", multiplied by current position.
Setting them equal kind of wrote me into a corner.
So this is a problem with two unknowns (the two angles of the hands). You need two different equations to solve it--in this case, one of the equations is simply A == B. The other is calculating A and B as an angle as a factor of time, and it turns out that B / 360 is the fraction of the hour that has gone by. So multiplying that fraction by the smaller 30 degrees that the hour hand travels in that same time gives us the number of degrees past the 7, then add the other 7 hours it's already traveled, and you have that relationship equation I posted.
Indeed! For opposite, B=A-180.
Gives A * 11/12=7 * 30-15
A=212.727272etc degrees.
So 7:05:27.2727...
Roughly 32min44s between the two occurrences, and that's true for every subsequent occurrence. This value can also be derived by considering there are 21 alignments between noon and midnight (not counting these), so dividing 720 minutes by 22 intervals