# In theory, this should be easy

1. May 29, 2008

### joeyar

In theory, this should be easy....

... During the space of 12 hours, how many times will the big hand of a clock form a right angle with the little hand?

2. May 29, 2008

### humanino

23 times ?

3. May 29, 2008

### joeyar

Close but no cigar.

4. May 30, 2008

### Jimmy Snyder

It depend upon when you start counting the 12 hours.

5. May 30, 2008

59x2--each click on the clock can have a right hand right angle and left handed one...

6. May 30, 2008

### humanino

Yes. If you start counting at 3:00 or 9:00 for 12:00+epsilon hours, with epsilon positive as small as you want, it seems it makes 23 times. If epsilon is negative, that probably makes only 22.

7. May 30, 2008

### phyzmatix

Does it have an hour, minute as well as second hand? Or only hours and minutes?

8. May 30, 2008

OOps--forget previous answer...thinking 24 now...

9. May 30, 2008

I was thinking second hand before

10. May 30, 2008

### joeyar

Completely ignore the seconds hand, we are only considering the minutes and hours hand. For argument's sake, start at the current time as I post this here in my timezone, which is 1.05.

But it seems to me that humanino has probably got it. :)

11. May 30, 2008

### humanino

Well, I got the first (3:00) yesterday before going to bed, and I stupidily was happy about myself, so I forgot the other one (9:00)

12. May 31, 2008

### Jimmy Snyder

There are others as well.

13. May 31, 2008

### Mokae

Yes, so difficult the problem

14. May 31, 2008

### Redbelly98

Staff Emeritus
I have used a "brute force" method, simply writing down the (approximate) times at which right-angles occur. Highlight the text below for the solution.

Also ... somebody may want to check if I missed any

There are 22 times where right-angles occur.

Note: times are approximate.

12:15
12:50
1:20
1:55
2:25
3:00
3:30
4:05
4:35
5:10
5:40
6:15
6:45
7:20
7:55
8:25
9:00
9:30
10:05
10:40
11:10
11:45

Last edited: May 31, 2008
15. May 31, 2008

### DavidWhitbeck

There is an analytic way to approach the problem--

The position vector of the big hand is given by (where t=0 is noon)

$$\mathbf{x_b}(t) = \sin \omega_b t \mathbf{i} + \cos \omega_b t \mathbf{j}$$

while the position vector of the little hand is given by

$$\mathbf{x_l}(t) = \sin \omega_l t \mathbf{i} + \cos \omega_l t \mathbf{j}$$

and you require that they are perpendicular to each other so that

$$\mathbf{x_b}(t)\cdot\mathbf{x_l}(t) = 0 \Rightarrow$$

$$\cos\omega_b t \cos\omega_l t + \sin \omega_b t \sin \omega_l t = 0 \Rightarrow$$

$$\cos (\omega_l - \omega_b)t = 0 \Rightarrow$$

$$(\omega_l - \omega_b)t_n = \pi n \Rightarrow$$

$$t_n = \frac{\pi n}{\omega_l - \omega_b}$$

but

$$\omega_b = 2\pi/T_b, \omega_l = 2\pi/T_l$$ where the T's are the periods.

So we have

$$t_n = \frac{n}{2} \frac{T_b T_l}{T_b - T_l}$$
where
$$T_b = 12 h, T_l = 1 h$$

16. May 31, 2008

### Kurdt

Staff Emeritus
Well its fairly easy to work out. angular velocity of each hand is simple.

Angular velocity of minute hand: $\omega_m=\frac{2\pi}{3600}$
Angular velocity of hour hand: $\omega_h = \frac{2\pi}{43200}$

Goverened by the rotational dynamics equations:

$$\theta = \theta_0 + \omega t$$

We want:

$$\theta_m - \theta_h = (2n+1) \frac{\pi}{2}$$

and we get:

$$t = \frac{(2n+1) \pi}{2(\omega_m - \omega_h)}$$

First couple of answers are (assuming you start from 12:00):

12:16:21.82
12:49:05.45
..etc

EDIT: I was wondering why my tex had come out as vectors for a second.

17. May 31, 2008

### DavidWhitbeck

Oh yeah woops I made a rookie mistake-- should have been (2n+1)/2 \pi as Kurt has it and not \pi n. Duh!

18. May 31, 2008

### joeyar

You have the right number. Well done.

Last edited: May 31, 2008
19. Jun 6, 2008

### phlegmy

wow someone brought maths into it.
i'd say for the first hour
when the minute hand is 15mins past the hour hand
for the last hour
when the minute hand is 15mins befor the hour hand

for the 10 hours in between
at 15mins before and after the hour hand

10*2 + 2*1 =22?

20. Jun 12, 2008

### Doctoress SD

Are we counting every second, minute or hour? LOL, J/K.