# Vector Clock Question

This is my first post here so please be gentle if I did anything wrong. Also this is for Emag class and its a 4000 level class but this question seems better suited for this forum. I have been doing homework all day and I do not knwo if I am just tired, but I cant get this.

## Homework Statement

Work this problem in both Cartesian (rectangular) coordinates and plane polar coordinates:

Consider an analog watch that keeps perfect time and assume the origin to be at the center of the dial, the x axis passing through the 12 mark, and the y axis passing through the 3 mark.

(a) Write the expression for the time-varying unit vector directed along the hour hand of the watch.
(b) Write the expression for the time-varying unit vector directed along the minute hand of the watch.
(c) Obtain the speciﬁc expression for these unit vectors when the hour hand and the minute hand are aligned
exactly and [are] between the 5 and 6 marks.

## Homework Equations

The hour hand makes jumps of 30 degrees or pi/6 and the minute hand jumps
6 degrees at each interval

the degrees have to be negative to move in a clockwise direction.

conversion from Cartesian to plane polar

## The Attempt at a Solution

I got the jump period and decide to arbitrarily make the scaler for minute 2 and the scaler for hours 1 I just dont understand how to make it move through the degrees in a time varying fashion.

all help greatly appreciated - I just need a clue.

Last edited by a moderator:

Ok so I am making progress. for anyone else who comes across this thread witha similar question this should help.

The angle R is calculated by (T/a)*360

where a is 30 degrees for hours and 6 degrees for minute

the final form is cos(R)x+sin(R)y

the real question comes down to the last part. still working on that

I bet you're in my class. Did you forget to make the 12 point in the positive X direction?

From there figure out how many degrees the hour hand sweeps per second and how many degrees the second hand sweeps per second. Then just subtract the time multiplied by degrees per second from the number of degrees in a circle.

As to the last part have fun with that. I used a brute force method, but I'm sure there is something elegant you could do.