After reviewing the statements more carefully, I have deviced a method to determine the "real length" of a pencil (stick). If I can determine the values of the y (length) and x (time) factors, it will be simple to determine the pencil's length. In order to determine the y & x values, the following information is assumed/given: $g_1 = g_2 = g = 9.8 m/s^2$ ; $c_1 = c_2 = c$
I, and everything in the room shrunk by y = 1/2.
My thumb width tw, defined as 2cm long.

Solution: find out the shrink factor and thereby be able to find the length of the pencil.

Method: create a "pendulum clock" who's period is 1 second (local time). The length of a string for such pendulum is given by $L_2 = \frac{g{t_2}^2}{4{\pi}^2}$. For $T_2$ = 1 second, $L_2 = 24.85cm$. Using my thumb, I measure the string and create a "pendulum clock". Keep in mind that since I am shrunk by 1/2, $L_2$'s "real" length ($L_1$) is $\frac{L_2}{2}$, or 12.42cm. So,outside my shrunk room, this pendulum would have a period of $T_1 = 2 \pi$ $(\frac{L_1}{g})^{1/2} = \frac{1}{2^{1/2}}sec$. Using a fiber optic cable of length (d) of 3 x 10^10; an LED; and a light detector; the time delay should be $t_2$ = d/c = 1sec. However, the "real time" $t_1$ is = d/2c = 1/2 sec (since the real length is d/2), but since I am measuring time with my pendulum, then the pendulum period ($T_2$) required is $\frac{1/2}{\frac{1}{2^{1/2}}}$ = $\frac{1}{2^{1/2}}$. What must I do now to the string length so the pendulum's period is actually 1/2 sec? I have to multiply its period by $\frac{1}{2^{1/2}}$, which means its length must be multiplied by 1/2, thus factor y = 1/2. So now, if you give me a stick that's 20cm long, I will determine it's 40cm long, but because I know the shrink factor is 1/2, I can say that the length of the stick is really 20cm.

After reviewing the statements more carefully, I have deviced a method to determine the "real length" of a pencil (stick). If I can determine the values of the y (length) and x (time) factors, it will be simple to determine the pencil's length. In order to determine the y & x values, the following information is assumed/given: $g_1 = g_2 = g = 9.8 m/s^2$ ; $c_1 = c_2 = c$
I, and everything in the room shrunk by y = 1/2.
My thumb width tw, defined as 2cm long.

Solution: find out the shrink factor and thereby be able to find the length of the pencil.

Method: create a "pendulum clock" who's period is 1 second (local time). The length of a string for such pendulum is given by $L_2 = \frac{g{t_2}^2}{4{\pi}^2}$. For $T_2$ = 1 second, $L_2 = 24.85cm$. Using my thumb, I measure the string and create a "pendulum clock". Keep in mind that since I am shrunk by 1/2, $L_2$'s "real" length ($L_1$) is $\frac{L_2}{2}$, or 12.42cm. So,outside my shrunk room, this pendulum would have a period of $T_1 = 2 \pi$ $(\frac{L_1}{g})^{1/2} = \frac{1}{2^{1/2}}sec$. Using a fiber optic cable of length (d) of 3 x 10^10; an LED; and a light detector; the time delay should be $t_2$ = d/c = 1sec. However, the "real time" $t_1$ is = d/2c = 1/2 sec (since the real length is d/2), but since I am measuring time with my pendulum, then the pendulum period ($T_2$) required is $\frac{1/2}{\frac{1}{2^{1/2}}}$ = $\frac{1}{2^{1/2}}$. What must I do now to the string length so the pendulum's period is $\frac{1}{2^{1/2}}$ sec (instead of 1 sec)? I have to multiply its period by $\frac{1}{2^{1/2}}$, which means its length must be multiplied by 1/2, thus factor y = 1/2. So now, if you give me a stick that's 20cm long, I will determine it's 40cm long, but because I know the shrink factor is 1/2, I can say that the length of the stick is really 20cm.

If the answer does not have to be too accurate, I would use the width of my thumb. Using this method, a "regular 7 inch" pencil, would be about 9 thumbs long. If you want the answer in cm, it would be about 18cm (my thumb is about 2cm). This method works in either "realm" since both the pencil and my thumb are shrunk by the same factor.

After reviewing the statements more carefully, I have deviced a method to determine the "real length" of a pencil (stick). If I can determine the values of the y (length) and x (time) factors, it will be simple to determine the pencil's length. In order to determine the y & x values, the following information is assumed/given: $g_1 = g_2 = g = 9.8 m/s^2$ ; $c_1 = c_2 = c$
I, and everything in the room shrunk by y = 1/2.
My thumb width tw, defined as 2cm long.

Solution: find out the shrink factor and thereby be able to find the length of the pencil.

Method: create a "pendulum clock" who's period is 1 second (local time). The length of a string for such pendulum is given by $L_2 = \frac{g{t_2}^2}{4{\pi}^2}$. For $T_2$ = 1 second, $L_2 = 24.85cm$. Using my thumb, I measure the string and create a "pendulum clock". Keep in mind that since I am shrunk by 1/2, $L_2$'s "real" length ($L_1$) is $\frac{L_2}{2}$, or 12.42cm. So,outside my shrunk room, this pendulum would have a period of $T_1 = 2 \pi$ $(\frac{L_1}{g})^{1/2} = \frac{1}{2^{1/2}}sec$. Using a fiber optic cable of length (d) of 3 x 10^10; an LED; and a light detector; the time delay should be $t_2$ = d/c = 1sec. However, the "real time" $t_1$ is = d/2c = 1/2 sec (since the real length is d/2), but since I am measuring time with my pendulum, then the pendulum period ($T_2$) required is $\frac{1/2}{\frac{1}{2^{1/2}}}$ = $\frac{1}{2^{1/2}}$. What must I do now to the string length so the pendulum's period is actually 1/2 sec? I have to multiply its period by $\frac{1}{2^{1/2}}$, which means its length must be multiplied by 1/2, thus factor y = 1/2. So now, if you give me a stick that's 20cm long, I will determine it's 40cm long, but because I know the shrink factor is 1/2, I can say that the length of the stick is really 20cm.