"Carpe Diem"
The clock face above has several divisions that are very clearly displayed; each of these represents a portion (fraction) of a 24 hour period, representing one full rotation of the Earth on its axis.
The largest of these divisions can be considered the clock itself as it represents exactly “1/2” of this 24-hour period (12/24).
Since the 12-hour clock is displayed here, the remainder of discussion below will be based on the assumption that one trip around the clock face will constitute one entire “unit”. High-noon will be considered that starting point; twelve o’clock midnight will represent the completion of the day’s “Post Meridiem” journey.
Ante Meridiem (A.M.): “before midday”
Post Meridiem (P.M.): “after midday”
Modular Arithmetic
When tracking time on a 12-hour clock, people are immersed in “Modular Arithmetic”; "mod 12" to be exact. On the 24-hour clock, 14 hours after midnight appears as “14:00”. On our 12-hour clock, this is referred to as “2:00 pm”, which is to say “2 hours after midday” (or 2 hours after 12-noon).
The generally accepted manner of interpreting modular arithmetic is summarized very briefly below.
14 – 2 = 12
Since the difference of 14 and 2 is a multiple of “12”, the statement below can be made.
14 Ξ 2 (mod 12)
15 – 3 = 12
Since the difference of 15 and 3 is also a multiple of “12”, the statement below can once again be made.
15 Ξ 3 (mod 12)
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Another very common example of modular arithmetic occurs when tracking days of the week. For example, October 1, 2011 falls on a Saturday; it is easily determined that October 29 also falls on a Saturday by looking at a Calendar. However, modular arithmetic, “mod 7” this time, can also be employed in coming to this conclusion, as shown below.
We know that October 1 falls on a Saturday; we would like to determine the day on which October 29 falls.
29 – 1 = 28
The difference of 29 and 1 is “28”, which is a multiple of “7”.
We can therefore make the following statement:
29 Ξ 1 (mod 7)
Since 29 Ξ 1 (mod 7), we can be sure that October 29 and October 1 fall on the same day of the week; more precisely they are exactly 4 weeks apart, since there are 4 multiples of 7 in 28 (28 being the difference between the two numbers being compared).
The day on which October 31 falls can be determined very easily as well. Since "31" is 2 days past "29", Halloween Day falls on Monday (2 days after Saturday).
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......back to our clock
One other consideration here will eliminate the need for a duplication of reasoning. Since seconds relate to minutes in the same way that minutes relate to hours, there is no need to describe both cases; what applies to one relationship can be directly transferred to the other by simply changing the parameters.
The relationship between minutes, hours and their representation on the clock face will be detailed below.
Fractions
The clock face shows two obvious sets of divisions, each of which represents a fraction (portion) of the entire face.
The smallest divisions each represent the passage of “one minute”; there are 60 of these in total on the clock face. Each interval of one minute can therefore be considered to be 1/60th of a full hour.
Because these divisions are so numerous, it is helpful to divide these 60 increments into sub-groups to more easily track time.
This is entirely similar to keeping a “tally” when counting large numbers of items in which four vertical lines followed by a fifth line, drawn diagonally through the previous four, produces a series of “five-bar gates”, each of which symbolizes sub-groups of five items.
On the clock face, there are 12 groupings of five-minute intervals, accounting for a total of sixty minutes on one full trip around the clock face. Each sub-group of five-minute intervals is delineated by a symbol of some sort; the symbol chosen represents the set of natural (counting) numbers.
This lends itself very nicely to tracking the number of hours past 12-noon as well as how many multiples of five-minute intervals the minute hand has travelled from its starting position at the top of the hour.
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We can credit the Ancient Babylonians for much of our present day time-keeping strategies. Because “60” can be broken down into many factors (it is highly composite), “fractional” components are very numerous and easy to come by.
This civilization divided the night and day into 12 hours each; hours were divided further into 60 one-minute intervals and each minute into 60 seconds.
Similarily, each solar year was approximately 360 days in length; this was, in turn, divided into 12 months to approximate the number of lunar cycles in one year.
It is for this ease of fractional representation that the Ancient Babylonians favored the “Base-60” number system.
In addition, the 360 degree connection to circles is directly related to the 360 day solar year approximation mentioned above.
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Design a Clock
Suppose for a moment that we are in charge of designing a clock that will help us track the passage of time.
The first clock we design has only a minute hand along with 60 divisions on its face, each representing a one-minute interval. One complete rotation of this minute-hand (360 degrees) represents the passing of 60 minutes, also known as one hour.
It is soon decided that this clock would function much more effectively if it could also keep track of the number of hours passed. Knowing that there are 12 hours in each of a day and night, it is decided to add a second set of partitions to the clock face, this time to represent the number of hours in each day/night cycle.
There are 12 increments in this second set of partitions, each occurring at five-minute intervals. An “hour-hand”, slightly shorter than the minute-hand to set it apart, is created and attached to a gear designed to make one complete rotation for every 12 rotations of the minute-hand. This will ensure that 720 minutes tracked by the minute-hand in one "day" (12*360 degrees) will result in exactly 360 degrees of rotation (12 hours) by the hour-hand. The gear ratio for this requirement must be 1:12.
Modern clocks employ a harmonic oscillator that maintains a constant frequency, resulting in higher precision time keeping; this serves the same purpose as does the flywheel in your vehicle’s engine.
We would no doubt have discovered the need for this (and invented it) had someone else not beat us to it.
One complete rotation of the minute-hand (360 degrees) results in an advancement of one hour, or “1/12th” of a full rotation on the clock face by the hour-hand. Therefore, each “one-hour increment” on the clock face represents 1/12thof 360 degrees or 30 degrees (1/12 * 360). Table 1 below summarizes this relationship; the table begins at 12-noon.
Table 1
.….360-Degree Rotations………….............………Time Interval……..
Minute-hand…... Hour-hand………………. Minutes........…..Hours (as time)
…..1……….……....1*1/12=1/12……………….60 min………..…60*1/60= 1 o’clock
…..2……….……....2*1/12=2/12…………..….120 min……….…120*1/60= 2 o’clock
…..3………....…....3*1/12=3/12…………...….180 min……….…180*1/60= 3 o’clock
…..4……….……....4*1/12=4/12………….…...240 min………..…240*1/60= 4 o’clock
…..5……….……....5*1/12=5/12………......….300 min………..…300*1/60= 5 o’clock
…..6……….……....6*1/12=6/12…………...….360 min…….....…360*1/60= 6 o’clock
…..7……….……....7*1/12=7/12…………...….420 min……..……420*1/60= 7 o’clock
…..8……….……....8*1/12=8/12……………....480 min……......…480*1/60= 8 o’clock
…..9……….…..…..9*1/12=9/12……………....540 min………..….540*1/60= 9 o’clock
….10………..…....10*1/12=10/12………..…....600 min………..…600*1/60= 10 o’clock
....11………...…....11*1/12=11/12………….….660 min………...…660*1/60= 11 o’clock
...12………..……..12*1/12=12/12………….....720 min………...…720*1/60= 12 o’clock
The graph below shows a geometric interpretation of the contents of Table 1.
Graph 1
The function shown in Graph 1 is defined below.
f(x) = 1/60(x)
The slope of 1/60 in this function indicates that in order to realize a change of 1 unit (hour) on the vertical axis, an increase of 60 units (minutes) must take effect on the horizontal axis. Several points satisfying this function are listed below.
Table 2
Point ……………………..Interpretation
A (150, 2.5)…………… 150 minutesà2:30 pm
B (360, 6)……………….360 minutesà6:00 pm
C (525, 8.75).……….….525 minutesà8:45 pm
D (615, 10.25)…….…...615 minutesà10:15 pm
E (720, 12)……………..720 minutesà12:00 midnight
A Cyclic Function
Upon completion of one 12-hour cycle, the process repeats itself for the A.M. time frame. The graph below illustrates the cyclic nature of time from the perspective of minutes and hours. This graph tracks the motion of the minute-hand (blue) and the hour-hand (red) for a period of 12 hours.
The blue function, representing motion of the minute-hand, shows one cycle for every one-hour interval; hence, there are 12 of these cycles represented in the illustration below.
Graph 2
The defining functions of the sinusoidal graphs illustrated in Graph 2 are shown below.
I’ve used the cosine function to represent the behavior of the clock hands as its path more accurately reflects their movement from 12-noon; this position corresponds to the starting point of the cosine functions illustrate above. Clockwise motion is defined here as the positive direction of travel on these cosine functions, again for its relative ease in transferring information from the clock face to the corresponding graphs.
h(x)=4cos(2π*(1/12)x)+7 à red
m(x)=6cos(2π x)+7 à blue
Significant points on the function “h(x)” are listed below.
Table 3
Point …………………Interpretation…………………Radians
A (0, 11)……………….12:00 noon……………..……….0 rad
B (3, 7).………..………3:00 pm……………….…….…π/2 rad
C (6, 3)………..……….6:00 pm………………..….… 2π/2 rad
D (9, 7)………..……….9:00 pm………………...…… 3π/2 rad
E (12, 11)…………......12:00 midnight……...…...…4π/2 rad
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Radian Measure is introduced in Table 3 above. This is a measurement of rotation angles and is based on the ratio of arc length travelled on a circle to its radius.
Radian measure becomes very useful when determining angular (rotational) velocities, etc. as it takes into account the radius that actually defines the arc itself; the longer the radius, the higher the rotational velocity at the end of that radius.
To put this in perspective, recall the merry-go-round that we all loved to ride on as kids. Things moved very slowly if you were standing near its center; for a bigger thrill, you would move further outward. This outward migration increased the radius, placing the thrill seeker (you) on a larger arc of travel. By doing this, you experienced the much higher rotational velocity that must naturally occur by covering a greater distance in the same amount of time as those "not so brave souls" who maintained a position closer to the point of rotation.
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The “x” components of the points listed above in Table 3 increase by “3” in each successive point; this represents intervals of 3 hours, which are significant “benchmarks” on the 12-hour clock. Those positions correspond to the “quarter-cycle” increments occurring on the sinusoidal function representing the hour-hand (shown in red) in Graph 2.
The equilibrium represents the point of rotation of the hour and minute-hands. The “y” components of each point simply correspond to the position of the tip of the hour-hand relative to that equilibrium.
Graph 3 below contains the first 6-hour span of the illustration shown in Graph 2; once again, the red function describes the position of the hour-hand. Since the time interval spans only 6 hours in this illustration, the function in blue, representing the minute-hand, shows only 6 cycles; the red function representing the hour-hand reveals only half of its full cycle. This scale provides for an improved visual interpretation of the interaction between the two functions.
Graph 3
h(x)=4cos(2π*(1/12)x)+7 à red
m(x)=6cos(2π x)+7 à blue
Once again, several significant points on the function “h(x)” are listed below. Maximum positions on the blue function representing the minute hand signify the "top of the hour". Hence, the "bottom of each hour" can be represented by vertical lines passing through the minimum values of the function in blue.
Table 4
Point ................Interpretation
A (0, 11)................12:00 noon
B (1, 10.46).............1:00 pm
C (2, 9)...................2:00 pm
D (3, 7)...................3:00 pm
E (4, 5)....................4:00 pm
F (5, 3.54)................5:00 pm
G (6, 3)....................6:00 pm
Angular Velocity
The angular (rotational) velocities at which the minute and hour-hands travel are obvious and are summarized below.
Table 5
Minute-hand……………………..……Hour-hand
1 rotation/hour……………….…….1/12 rotations/hour
2π radians/hour……….……………2π/12 = π/6 radians/hour
12 rotations/day………….………….1 rotation/day
24π radians/day…………….……..2π radians/day
Instantaneous Velocity
Angular velocity remains constant for the hands of our clock (assuming an accurate clock).
Now suppose we are interested in determining how fast the hands of the clock are moving vertically from the top of the hour (12-noon) to the bottom of the clock (the 6 o’clock position).
Obviously, the minute-hand gets there faster than the hour hand, but its vertical rate of decent is variable. When it first moves from the 12-o’clock position, its vertical rate of change is very slow.
As it advances further along towards the “quarter-hour” mark (15 minute position), it accelerates from the “vertical rate of change” perspective.
It is easy to see that the minute-hand experiences its highest “vertical rate of change” at the exact moment it passes the “15-minute” mark. At this point, all rotational motion is being directed vertically downward; this is referred to as the “point of inflection” of the corresponding sinusoidal function. This rate of change is referred to as instantaneous velocity is can be easily determined by evaluating the first derivative of the function that corresponds to the minute-hand.
Refer to Graph 4 below for a visual representation of function m(x), measured over 1 hour.
Graph 4
m(x) =6 cos (2π x) + 7
m'(x) =-6 sin (2π x)* 2π
m'(x) =-12π sin (2π x)
Evaluating m'(x) =-12π sin (2π x) for x=0.25 hours results in the instantaneous rate of “vertical change” of the clock’s minute hand, defined by the function m(x) =6 cos (2π x) + 7 .
Several results of this analysis are shown in Table 5 below; the points identified are taken from Graph 4. I will assume the minute hand has a length of 6 cm.
Table 5
Point……………First Derivative……………Rate of Vertical Change
A (0, 13)………..m'(0) = 0……….…………...….......….0 cm/minute
B (1/4, 7)……..…m'(1/4) = -37.6991…..……..…-37.6991 cm/minute
C (1/2, 1)……..…m'(1/2) = 0…………..……....……..….0 cm/minute
D (3/4, 7)……..…m'(3/4) = +37.6991………..…+37.6991 cm/minute
E (4/4, 7)…………m'(1) = 0………………………..……..0 cm/minute
Graph 5 below illustrates the hour-hand’s journey over the course of a 12 hour period (1 “day”).
Graph 5
h(x) =4 cos (2π x) + 7
h'(x) =-4 sin (2π x)* 2π
h'(x) =-8π sin (2π x)
Evaluating h'(x) =-8π sin (2π x) for x=0.25 hours results in the instantaneous rate of “vertical change” of the clock’s hour-hand, defined by the function h(x) =4 cos (2π x) + 7.
Several results of this analysis are shown in Table 6 below; the points identified are taken from Graph 5. I will assume the hour-hand has a length of 4 cm and that 12 hours constitutes a “day”.
Table 6
Point……………First Derivative……………Rate of Vertical Change
A (0, 11)………...h'(0) = 0……….………….......…..….0 cm/day
B (1/4, 7)……...…h'(1/4) = -25.1327………...…-25.1327 cm/day
C (1/2, 3)…………h'(1/2) = 0…………..……....…....….0 cm/day
D (3/4, 7)……....…h'(3/4) = +25.1327…………+25.1327 cm/day
E (4/4, 11)……...…h'(1) = 0…………………...…..……..0cm/day
Comparing the values of the instantaneous rates of change of the two functions above can add some additional meaning to this entire scenario. We will quickly analyze the values of the first derivatives of each function at the “quarter cycle” below.
Table 7
Minute-hand……….Hour-hand………..Ratio of Velocities
-37.6991…………..……-25.1327……..…-37.6991:-25.1327 = 3:2
This ratio, “3:2 “, of corresponding instantaneous velocities at this point in time, is directly attributable to the ratio of the lengths of the minute and hour-hands on the clock. The longer the radius (hand), the higher the relative instantaneous velocity will be.
There is another interesting ratio that should be mentioned here before drawing the discussion to a close. If Graph 4 and Graph 5 had been drawn on the same scale, the two velocities shown in Table 7 would have differed by an additional factor of “12”, since that is the number of "minute-hand" cycles in a 12-hour period.
This can be verified by re-visiting the functions shown in Graph 2, finding the derivatives and evaluating those at the "quarter cycle" marks.
The End
