"The Hour Hand and the Minute Hand"

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zenith20

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The Hour Hand and the Minute Hand
We all know that the hour hand and the minute hand on a clock travel at different speeds. However, there are certain occasions when they are exactly opposite each other. Can you give a simple formula for calculating the times of these occasions?
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solution:
Here is the formula that gives the minutes past twelve to which the hour hand points when the minute hand is exactly thirty minutes ahead.
Minutes past twelve Y = 30/11 [(n-1) 2+1]
where n is the next hour
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could anyone help me understand how this formula is derived?

thank you in advance
 
Hello, zenith20!

I don't understand their formula either . . .

The Hour Hand and the Minute Hand

We all know that the hour hand and the minute hand on a clock travel at different speeds.
However, there are certain occasions when they are exactly opposite each other.
Can you give a simple formula for calculating the times of these occasions?

Solution:
Here is the formula that gives the minutes past twelve to which the hour hand points
. . when the minute hand is exactly thirty minutes ahead.

Minutes past twelve Y = 30/11 [(n-1) 2+1], .where n is the next hour. . ??
Click to expand...

\(\displaystyle \text{Here's the way I worked it out . . .}\)


\(\displaystyle \text{The minute hand moves }360^o\text{ in 60 minutes . . . It moves }6^o\text{ per minute.}\)
. . \(\displaystyle \text{In }t\text{ minutes, it moves }6t\text{ degrees.}\)

\(\displaystyle \text{The hour hand moves }30^o\text{ in 60 minutes . . . It moves }\tfrac{1}{2}^o\text{ per minute.}\)
. . \(\displaystyle \text{In }t\text{ minutes, it moves } \tfrac{1}{2}t\text{ degrees.}\)


\(\displaystyle \text{Suppose the hour hand is at }h\text{ o'clock.}\)
\(\displaystyle \text{It has already moved }30h\text{ degrees.}\)
\(\displaystyle \text{In the next }t\text{ minutes, it moves }\tfrac{1}{2}t\text{ degrees.}\)
\(\displaystyle \text{So its position will be: }\:H \:=\:30h + \tfrac{1}{2}t\text{ degrees.}\)

\(\displaystyle \text{In the same }t\text{ minutes, the minute hand has moved }6t\text{ degrees.}\)
\(\displaystyle \text{So we have: }\:M \:=\:6t\)


\(\displaystyle \text{When will }M\text{ be }180^o\text{ beyond }H?\)

\(\displaystyle \text{We have: }\:6t \;=\;30h + \tfrac{1}{2}t + 180\)

. . \(\displaystyle \text{which simplifies to: }\;\boxed{t \:=\:\tfrac{60}{11}(h + 6)}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

\(\displaystyle \text{We can check our formula, hour by hour.}\)
. . \(\displaystyle \text{(Note: at 12 o'clock, }h = 0.)\)

. . \(\displaystyle \begin{array}{ccc} \text{Hour} & \text{Minutes} & \text{Time} \\ \hline \\[-3mm] 12\!:\!00 & \frac{360}{11} \:=\:32\frac{8}{11} & 12\!:\!32\frac{8}{11} \\ \\[-3mm]1\!:\!00 & \frac{420}{11} \:=\:38\frac{2}{11} & 1\!:\!38\frac{2}{11} \\ \\[-3mm] 2\!:\!00 & \frac{480}{11} \:=\:43\frac{7}{11} & 2\!:\!43\frac{7}{11} \\ \\[-3mm] 3\!:\!00 & \frac{540}{11} \:=\:49\frac{1}{11} & 3\!:\!49\frac{1}{11} \\ \\[-3mm] 4\!:\!00 & \frac{600}{11}\:=\:54\frac{6}{11} & 4\!:\!54\frac{6}{11} \\ \\[-3mm] 5\!:\!00 & \frac{660}{11} \:=\:60\;\;\;\; & 5\!:\!60 \\ \vdots & \vdots & \vdots \end{array}\)


\(\displaystyle H\!A!\)

\(\displaystyle \text{The last one, }5\!:\!60\text{, means "60 minutes after 5 o'clock"}\)

. . \(\displaystyle \text{which, of course, is "6 o'clock".}\)

 
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