Overlapping Clock Hands and Tax Free Gifts
Tax Free Money
Suppose that you wanted to give someone a tax free gift of $1,000. How much do you have to give them if the tax rate is 20% ?
A quick calculation would add 20% of $1,000 or $200 to the amount, and give $1,200. But 20% of $1,200 is $240, so the recipient gets only $960 instead of $1,000 after paying the tax. The recipient gets $40 less than intended, so adding $40 for a total of $1,240 would seem to be the remedy. But the tax on this is $248 leaving the recipient $8 short.
This is an iterative problem which requires repeating the above procedure many more times to calculate the required total so that the recipient nets $1,000 after taxes. The sequence, a geometric infinite series, is as follows:
where S is the required sum, and n approaches infinity. The sum increases as shown below:
The iterations in [2] ending with 0.32 totals to 1249.92. More iterations show that the required total is approaching 1250. There is an easier way.
Factor out 1000 in [1] gives:
Multiply both sides of [3] by .2:
Subtract [4] from [3] gives:
Since n+1 is approaching infinity, the limit of .2 raised to that power equals 0. Thus:
20% of $1250 is $250, leaving $1000 after taxes.
Compare this to [1] with a=1000 and r=.2 or 20% or 1/5.
Factor (a) from [9]:
Multiply both sides by r :
Subtract [11] from [10]:
Solving for S:
Since n+1 is approaching infinity, the limit of r equals 0, but only if r is between -1 and 1, otherwise the limit does not equal zero and the series does not converge. Thus:
Substituting a=1000, and r=.2 as in the tax free money problem above gives:
What does this have to do with overlapping clock hands?
At what times do the minute hand and hour hand of a precise analog clock overlap? There are multiple ways to calculate this and are shown here and here and here. However, here are two more methods.
Method 1 will use a geometric infinite series as described above, and Method 2 will use a minimum/maximum method of differential calculus.
Figure 1
The minute hand is used to determine the position of the hour hand more precisely. The percent of rotation of the minute hand indicates the percent of movement of the hour hand between the hour marks.
The minute hand rotates at 360°/hour; the hour hand 30°/hour. The rotation of the hands are not independent. The rotation of one depends on the rotation of the other. My definition of a precise clock has nothing to with the exact time, but with the correct movement of the hands relative to each other. The minute hand will never point to 6 while the hour hand points directly at 1 or 2 or any hour. It will be halfway between the hours. When the minute hand is at 9, the hour hand must be 3/4 of the way between hours. See the figures above.
The minute and hour hand will overlap when their angles with respect to noon or midnight are equal. For this discussion, a positive angular rotation will be in the clockwise direction. At noon or midnight, the angle of the hour hand will be 0°, 1:00 will be 30°, 2:00 will be 60°, and so on. In each of these cases, the angle of the minute hand will be 0°. The angle in degrees of each hand as a function of time is shown below:
where m is the angle in degrees of the minute hand,
h is the angle in degrees of the hour hand, and
t is the time in decimal hours.
In decimal hours, 1:00 o'clock is 1.00, 1:15 is 1.25, 1:30 is 1.50, etc. Times will be shown in multiple formats as necessary.
The first (and obvious case) of overlap is at noon, 12:00:00, or t=0. From [16], set m = h:
Solving, t=0, or both hands are pointing up and the angle of each is 0°.
The next case is after 1:00. 1:05 seems like a reasonable choice, but as the minute hand approaches the hour hand, the hour hand is moving away at 1/12 the rotational speed. So an overlap will happen, but where?
At 1:00, the hour hand is pointing at the 1 position at 30°; the minute hand is at 0°. Let the angles of h and m equal:
Solving, t = 1/11 hours after 1:00, or 1 1/11 hours, or 12/11 hours or exactly 1:05:27 3/11 o'clock or 5 minutes 27 and 3/11 seconds after 1:00 0'clock. (To convert from decimal hours to hours. minutes, and seconds, see here. Also most scientific calculators can convert to DMS (degrees minutes seconds) and back.)
At 1:05, the minute hand moves to 30° and the hour hand, that was at 30° at 1:00, has moved an additional 30/12 degrees because the hour hand is moving 1/12 the rate of the minute hand -- 30 degrees per hour vs 360 degrees per hour.
Solving, t=1/132 hours after 1:05 (13/12 hours) equals 12/11 hours. Or again, exactly 1:05:27 3/11 o'clock as above.
Because the hour hand is rotating at 1/12 the rate of the minute hand, the 1/12 can be treated as "r" in [9] with "a" as any hour at the zero position. The sequence is also similar to [1] and it's derivation. "a" will be 1 to show that the overlap time agrees with the above two calculations. "t" is being used for S.
A quick calculation would add 20% of $1,000 or $200 to the amount, and give $1,200. But 20% of $1,200 is $240, so the recipient gets only $960 instead of $1,000 after paying the tax. The recipient gets $40 less than intended, so adding $40 for a total of $1,240 would seem to be the remedy. But the tax on this is $248 leaving the recipient $8 short.
This is an iterative problem which requires repeating the above procedure many more times to calculate the required total so that the recipient nets $1,000 after taxes. The sequence, a geometric infinite series, is as follows:
where S is the required sum, and n approaches infinity. The sum increases as shown below:
The iterations in [2] ending with 0.32 totals to 1249.92. More iterations show that the required total is approaching 1250. There is an easier way.
Factor out 1000 in [1] gives:
Multiply both sides of [3] by .2:
Subtract [4] from [3] gives:
Since n+1 is approaching infinity, the limit of .2 raised to that power equals 0. Thus:
20% of $1250 is $250, leaving $1000 after taxes.
Geometric Infinite Series
The general form for a geometric infinite series is:
Compare this to [1] with a=1000 and r=.2 or 20% or 1/5.
Factor (a) from [9]:
Multiply both sides by r :
Subtract [11] from [10]:
Solving for S:
Since n+1 is approaching infinity, the limit of r equals 0, but only if r is between -1 and 1, otherwise the limit does not equal zero and the series does not converge. Thus:
Substituting a=1000, and r=.2 as in the tax free money problem above gives:
What does this have to do with overlapping clock hands?
Overlapping Clock Hands
Method 1 will use a geometric infinite series as described above, and Method 2 will use a minimum/maximum method of differential calculus.
Method 1 - Infinite Geometric Series
With an analog clock, time is displayed by the hour hand. When the hour hand points at exactly 1, it is 1:00. When it is exactly half-way between 1 and 2, it is 1:30. When it is 3/4 of the way between 1 and 2, it is 1:45.Figure 1
The minute hand is used to determine the position of the hour hand more precisely. The percent of rotation of the minute hand indicates the percent of movement of the hour hand between the hour marks.
The minute hand rotates at 360°/hour; the hour hand 30°/hour. The rotation of the hands are not independent. The rotation of one depends on the rotation of the other. My definition of a precise clock has nothing to with the exact time, but with the correct movement of the hands relative to each other. The minute hand will never point to 6 while the hour hand points directly at 1 or 2 or any hour. It will be halfway between the hours. When the minute hand is at 9, the hour hand must be 3/4 of the way between hours. See the figures above.
The minute and hour hand will overlap when their angles with respect to noon or midnight are equal. For this discussion, a positive angular rotation will be in the clockwise direction. At noon or midnight, the angle of the hour hand will be 0°, 1:00 will be 30°, 2:00 will be 60°, and so on. In each of these cases, the angle of the minute hand will be 0°. The angle in degrees of each hand as a function of time is shown below:
where m is the angle in degrees of the minute hand,
h is the angle in degrees of the hour hand, and
t is the time in decimal hours.
In decimal hours, 1:00 o'clock is 1.00, 1:15 is 1.25, 1:30 is 1.50, etc. Times will be shown in multiple formats as necessary.
The first (and obvious case) of overlap is at noon, 12:00:00, or t=0. From [16], set m = h:
Solving, t=0, or both hands are pointing up and the angle of each is 0°.
The next case is after 1:00. 1:05 seems like a reasonable choice, but as the minute hand approaches the hour hand, the hour hand is moving away at 1/12 the rotational speed. So an overlap will happen, but where?
At 1:00, the hour hand is pointing at the 1 position at 30°; the minute hand is at 0°. Let the angles of h and m equal:
Solving, t = 1/11 hours after 1:00, or 1 1/11 hours, or 12/11 hours or exactly 1:05:27 3/11 o'clock or 5 minutes 27 and 3/11 seconds after 1:00 0'clock. (To convert from decimal hours to hours. minutes, and seconds, see here. Also most scientific calculators can convert to DMS (degrees minutes seconds) and back.)
At 1:05, the minute hand moves to 30° and the hour hand, that was at 30° at 1:00, has moved an additional 30/12 degrees because the hour hand is moving 1/12 the rate of the minute hand -- 30 degrees per hour vs 360 degrees per hour.
Solving, t=1/132 hours after 1:05 (13/12 hours) equals 12/11 hours. Or again, exactly 1:05:27 3/11 o'clock as above.
Because the hour hand is rotating at 1/12 the rate of the minute hand, the 1/12 can be treated as "r" in [9] with "a" as any hour at the zero position. The sequence is also similar to [1] and it's derivation. "a" will be 1 to show that the overlap time agrees with the above two calculations. "t" is being used for S.
Using [14] to solve the series above gives:
Thus, t=12/11 hours, or 1 1/11 hours, or 1:05:27 3/11 o'clock as above.
Using [12], and varying "a" will give the following exact times when the minute hand and the hour hand overlap:
Table 1
Note that each overlap is separated by exactly 12/11 hours.
Method 2 - Min/Max through Differentiation
Assume that we have a clock with the minute hand and hour hand of equal length of one unit.
Figure 2
Figure 2 shows a portion of a clock face as follows:
- Point A - the center of the clock face.
- Line segment AB - the minute hand of length 1 unit pointing at 12:00.
- Line segment AC - the hour hand of length 1 unit pointing at 3:00.
- Line segment BC - the distance between the ends of the minute and hour hand.
Figure 2 shows a portion of a clock face as follows:
- Point A - the center of the clock face.
- Line segment AB - the minute hand of length 1 unit pointing at 12:00.
- Line segment AC - the hour hand of length 1 unit pointing at 3:00.
- Line segment BC - the distance between the ends of the minute and hour hand.
- The length of BC in figure 2 can be determined using the cosine rule:
Radians will be used for angular measure:
Radians will be used for angular measure:
As in [16] but with the angle now in radians, the angle of each hand as a function of time in hours is shown below:
The angle in radians between the hands, A in figure 2, as a function of time is:
The distance squared between the ends of the hands using [22] is:
To find the times when the distance (BC) is 0, take the derivative of [27]. The distance [BC] will be 0 when the square of [BC] is 0, which simplifies the calculations.
To find the times when the distance (BC) squared is at a minimum, set [29] equal to 0 and solve for t. Setting [29]=0 is not solving for when the ends of hands are 0 units apart. It is solving for when they are at a minimum distance, which happens to be 0 because of their length. Also note that this will determine the maximum distance between the ends of the hands because setting [29]=0 determines when the slope of [27] is 0, at both maximums or minimums if they exist.
The general solutions for [30] are:
where n is any integer, and t is time in decimal hours.
Sequencing n in [31A] gives the same results as shown in Table 1.
Sequencing n in [31B] gives the times when the ends of the hands are the farthest apart or in a straight line. When n=5 here, the time is 6:00.
If you find any errors or have any suggestions, please contact me.
[QED]
John L. Ferri
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