JPS6156110B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION In recent years, printers that express characters by an array of dots, that is, dot matrix printers, have become widely used. In particular, dot-impact printers, which print dots by mechanically striking the tip of a needle onto printing paper via an ink ribbon, have become widely used as office printers because they can print copies at the same time.

Recently, dot matrix type line printers have appeared and are becoming widespread. A dot matrix type line printer has a single hammer that prints a dot every character or several characters, and the hammer is swung horizontally by the width of the character being supported, while at the same time moving the printing paper vertically. The hammer is actuated at a predetermined position to form letters in the form of dots.

FIG. 1 is a diagram for explaining the process of forming characters according to the procedure described above. Initially, the printing hammer is in position 1 and moves laterally towards position 2.
During this time, the printing paper is stopped. When the printing hammer reaches position 2, the printing paper moves upward by one pitch, and the printing hammer comes to be located at position 3. Next, the printing hammer moves to the left and reaches 4.
Then the printing paper moves upwards by one pitch again,
The printing hammer will now be located at position 5.
Repeating this operation, the printing hammer changes from 5 to 6.
→Move to 7→8→...14→. In the meantime, the printing hammer is at a predetermined position a where printing is to be performed,
By activating the printing hammer and printing the dots at the moment when b, c, ... z, ... a', b' are reached,
Desired characters can be formed. When a character is formed by seven rows of dots in the vertical direction as shown in FIG. 1, the character is formed during 3.5 reciprocations of the printing hammer.

Further, a large number of printing hammers are arranged in the horizontal direction, and these printing hammers are integrated to form a hammer bank, and this hammer bank swings from side to side. One line of printing is completed during 3.5 reciprocations of this hammer bank.

The hammer bank does not need to swing at a constant speed, nor does it need to remain stationary while the printing paper moves one pitch. For example, a character is formed by oscillating a hammer bank with a sine wave having an amplitude slightly larger than the width of two characters, providing a position sensor on the hammer bank, and operating a printing hammer at a predetermined position. be able to. Normally, line printers print 132 characters per line. Therefore, if one hammer is to print two characters as shown in FIG. 1, it is necessary to provide 66 printing hammers. Therefore, even if considerable effort is made to reduce the weight, the weight of the hammer bank will be close to 1 kg.

As shown in Figure 1, 2
In order to print characters, it is necessary to swing the hammer bank with a swing width of at least about 5 mm. When swinging with a sine wave as described above, it is necessary to swing with a width of about 7 mm with some allowance.

Furthermore, in order to achieve a printing speed of 600 lines/minute, it is necessary to print at a speed of 10 lines per second. On the other hand, as mentioned above, it is necessary to make 3.5 reciprocations of the hammer bank to form one line character. Furthermore, when considering the spacing between lines, it is necessary to make the hammer bank reciprocate at least four times in order to print one line. Therefore, the swinging frequency of the hammer bank is required to be as high as 40 times/second.

In other words, in a dot matrix type line printer, a load of about 1 kg is applied at 40 Hz with a swing width of about 7 mm.
must be oscillated at a frequency of When such rocking is performed, a considerably large vibration force acts on the housing of the device, which poses a practical problem.

For example, when a weight of 1 kg as described above vibrates with a 40 Hz sine wave with an amplitude of 7 mm, the maximum value of the vibration force acting on the housing reaches 23 kg.

Dot matrix type line printers are generally made to be small and lightweight, so when a large vibration force of 23 kg is applied, the entire device shakes greatly and at the same time generates a large amount of noise, resulting in a serious defect.

Also, as mentioned above, the weight is 1kg, the swing width is 7mm, and the frequency is 40mm.
To swing at Hz, the peak driving force is 23Kg.
This requires the use of a fairly large drive source, making the cost of the drive source and power source excessive.
It also becomes an obstacle to miniaturization and weight reduction.

Generally, in order to remove the vibration force caused by the swinging of an object, a balancing weight having the same weight as the swinging object is provided, and this balancing weight is always swung in the opposite direction to the swinging object. is taken. In this method, the vibration force can certainly be removed, but the driving force is doubled. Therefore, although the problem of vibration force is solved, the problem of driving force becomes even worse, and this cannot be said to be a desirable solution.

An object of the present invention is to provide a method for simultaneously solving the problems of vibration force and driving force as described above. That is, the present invention provides a hammer bank swinging mechanism that can reduce the vibration force generated when swinging the hammer bank in a dot matrix type line printer and at the same time reduce the driving force required to swing the hammer bank. is intended to provide.

FIG. 2 is a diagram for explaining the operating principle of the hammer bank swinging mechanism of the present invention. In the figure, 21 is a hammer bank, 22 is a weight for balance, 23 and 24 are leaf springs, 25 is a link,
26 is a pin provided on the hammer bank 21;
7 is a pin provided on the weight 22, 28 is a link 2
5 fulcrum points are shown. Furthermore, X ₁ and X ₂ represent the displacements of the hammer bang 721 and the weight 22 by ₁ and _2, respectively.
represent the driving forces to be applied to the hammer bank 21 and the weight 22, respectively.

As shown in FIG. 2, the hammer bank swinging mechanism of the present invention has a balance weight 22, and the hammer bank 21 is connected to the weight 22 by a link 25. Therefore, the hammer bank 21 and the weight 22 swing in opposite directions at a swing width ratio determined by the length of the arm of the link 25.

The hammer bank 21 and the weight 22 are connected by two leaf springs 23. Further, the weight 22 is fixed to the base via two leaf springs 24. Here, the weight of the hammer bank 21 is w ₁ (Kg), and the weight 2
2 weight w ₂ (Kg), spring constant k ₁ (Kg/
mm), the spring constant of the leaf spring 24 is k ₂ (Kg/mm), and the lengths of the link arms (see Figure 2) are l ₁ (mm) and l ₂ (mm).

Now, when the hammer bank 21 is oscillated by x ₁ =x·sinωt, the weight 22 is oscillated by x ₂ =−l ₂ /l ₁ ·xsinωt.

However, ω is the angular velocity (rad/sec), and t is the time (sec). Then, the forces ₁ and ₂ required to drive the hammer bank 21 and the weight 22 are calculated using the following formula. ₁ =w ₁ /g・d ² x ₁ /d+ ₂ +k ₁・(x ₁ −x ₂ )=
{−w/g・ω ₂ +k ₁・(1+l ₂ /l ₁ )}・x・sinωt ...(1) ₂ = w ₂ /g・d ² x ₁ /d+ ₂ +k ₁・(x ₂ −x ₁ )＋
k ₂・x ₂ = {l ₂ /l ₁・w ₂ /g・ω ² −k ₁・(1+l ₂ /l ₁
) −l ₂ /l ₁・k ₂ }・x・sinωt ...(2) ₁ and ₂ are shown in FIG. 2 to act on mutually opposite sides of the fulcrum 28 of the link 25,
Now, assuming that a driving source (not shown) is connected and driven only to the hammer bank 21, the force m required for the driving source is m= ₁ -l ₂ /l ₁・₂ ...(3) Become. Calculating this m from equations (1) and (2), m={k ₁・(1+l ₁ /l ₁ ) ² +k ²・(l ₂ /l ₁
) −(w ₁ /g+l ₂ ² /l ₁ ² w ₂ /g)・ω ² }
・x・sinωt...(4)

On the other hand, the reaction force _Rn received from the drive source on the base on which the drive source is attached _{is Rn} = ₁ + ₂ = {(l ₁ /l ₁・w ₂ /g−w ₁ /
g)・ω ² −l ₂ /l ₁・k ₂ }・x・sinωt……(5) Furthermore, the reaction force _Rs that the base receives from the leaf spring 24 is _RS = k ₂・x ₂ = −l ₂ /l ₁・k ₂・x・sinωt……(
6) The force currently acting on the base, i.e. the reaction force _Rn from the driving source
If the reaction force _Rs received from the plate spring 24 is equal, no vibration force will act on the base. That is, {(l ₂ /l ₁・w ₂ /g−w ₁ /g)・ω ² l ₂ /l
₁・k ₂ }・x・sinωt = −l ₂ /l ₁・k ₁・x・sinωt ……(7) If you set the constants of each part so that it holds true, you can create a quiet hammer without applying vibration force to the base A rocking motion of the bank can be realized.

In order for equation (7) to hold, l ₂ /l ₁ ·w ₂ /g w ₁ /g=0, that is, w ₁ /w ₂ =l ₂ /l ₁ ...(8) should hold. That is, by determining the lengths of the hammer bank 21, the weight 22, and the arm of the link so that equation (8) holds true, quiet rocking of the hammer bank can be achieved.

On the other hand, when each constant is determined so that equation (8) holds true, the driving force of equation (4) becomes as follows.

m={k ₁ (1+l ₂ /l ₁ ) ² +k ₂ (l ₂ /l ₁ ) ² −w ₁ /g(1+l ₂ /l ₁ )ω ² }xsinωt...(9
) Now ω ² = {k ₁ (1+l ₂ /l ₁ ) ² +k ² (l ₂ /l ₁ ) ²
}/ {w ₁ /g(1+l ₂ /l ₁ )} ...(10) When ω is substituted into equation (9), m=0.

In other words, this means that the driving force m becomes essential if the vibration is made at a frequency that satisfies equation (10).

The above results show that it is possible to realize a hammer bank swinging mechanism that can be driven with a very small driving force without applying any vibration force to the base or the housing.

That is, the weight of the hammer bank 21 w ₁ the weight of the balance weight 22 so as to satisfy equations (8) and (10)
By determining the relationship between w ₂ , the arm lengths l ₁ and l ₂ of the link 25, the swing frequency ω, and the spring constants k ₁ and k ₂ of the leaf springs 23 and 24, no vibration force is generated and the driving force is It is possible to realize a hammer bank swinging mechanism that is almost unnecessary.

Now _, in the hammer _bank swinging mechanism based on the principle _shown in FIG. constant
The case where k ₂ is taken to be equal will be explained.

In this case, since w ₁ = w ₂ , we must set l ₁ = l ₂ from equation (8). Also, since k ₁ and k ₂ , equation (10) becomes as follows.

ω ² =5/2・g・k ₁ /w ₁ ...(11) Here, w ₁ is determined by the printing hammer calculation, and ω is determined from the printing speed, so from equation (11), k ₁ can be determined.

That is, k ₁ =2/5·w ₁ /g·ω ² (12) In this way, the constants of each part of the hammer bank swinging mechanism based on the principle shown in FIG. 2 can be determined.

Note that it is not always necessary to make the weight w ₂ of the balance weight 22 equal to the weight w ₁ of the hammer bank 21.

If the weight w ₂ of the balance weight 22 is larger than the weight w ₁ of the hammer bank 21, then l ₂ becomes l ₁ from equation (8).
Must be made smaller. Therefore, the swing amplitude of the balance weight 22 can be made smaller than the swing amplitude of the hammer bank 21.

Furthermore, if the weight w ₂ of the balance weight 22 is smaller than the weight w ₁ of the hammer bank 21, l ₂ must be larger than l ₁ from equation (8). Therefore, the amplitude of the swing of the balance weight 22 is larger than the amplitude of the swing of the hammer bank 21. From equation (8), when ω ₂ approaches zero, l ₂ /l ₁ approaches infinity. The amplitude of the swing of the hammer bank 21 is determined by the size of the printed characters and the number of characters that one printing hammer handles, so if _w2 is made very small, the width of the swing of the balance weight 22 will be It becomes very large and impractical.

On the other hand, the spring constant k ₁ of the leaf spring 23 and the spring constant k ₂ of the leaf spring 24 do not need to be equal. In equation (10), the swing frequency ω, the weight w ₁ of the hammer bank 21, and the link 2
Assuming that the arm length ratio l ₂ /l ₁ of 5 is determined, k ₁・(1+l ₂ /l ₁ ) ² +k ₂・(l ₂ /l ₁ ) ² =w ₁ /g・( 1+l ₂ /l ₁ )・ω ² ...(13) k ₁ and k ₂ must be determined so as to satisfy the following. Therefore, if k ₂ is increased, k ₁ is decreased, and k ₂
If you decrease , you need to increase k ₁ .

The problem with the mechanism of the present invention is the driving force at the time of startup. The driving force m at any oscillation frequency is
It is expressed by equation (9) and becomes zero at the frequency expressed by equation (10). However, at startup, the value becomes quite large.
That is, when ω=0 and ωt=π/2 in equation (9), m={k ₁ (1+l ₂ /l ₁ ) ² +k ₂ (l ₂ /l ₁ ) ² }x (14) becomes. Now, if l ₂ /l ₁ = 1 k ₁ = k ₂ , then m
= 5k ₁ x, which is a fairly large value. Therefore, although the driving frequency expressed by equation (10) eliminates the need for starting force, a large force is required at the time of starting, creating a paradox in which a small driving source cannot be used.

In the present invention, the power transmission mechanism is devised in order to prevent the drawback that a large force is required at startup as described above.

FIG. 3 is a diagram showing an embodiment of the present invention.
Reference numeral 28 indicates the same member as the member with the same number in FIG. Reference numeral 29 indicates a connecting rod of the crank, and 30 indicates a crank mechanism for reciprocating the crank connecting rod.

Now, if the length of the crank connecting rod is made long enough, the hammer bank 21 can be rocked almost in a sinusoidal motion. Therefore, if the swinging motion of the hammer bank 21 matches the frequency of equation (10), almost no load will be applied to the crank mechanism 30. However, since the frequency is zero at startup, a large load is placed on the crank mechanism 30 as described above. In order to avoid this load at startup, a special mechanism is adopted in the crank mechanism 30 shown in FIG. 3. In other words, when the rotation speed is low, the length of the crank arm is zero, and as the rotation speed increases, the centrifugal force acting on the weight causes the crank arm to reach a predetermined length. A mechanism is provided in the crank mechanism 30.

4 and 5 show the above-mentioned crank mechanism 30.
FIG. 4A and FIG. 5A are detailed views of FIG. 4A and FIG.
3 shows the view from the direction of arrow A in FIG.
Further, FIG. 4C is a sectional view taken along the line CC in FIG. 4A. Furthermore, Figure 5B is D-D of Figure 5A.
A cross-sectional view along the line is shown. In addition, Fig. 4 A, B,
C shows a state where the rotational speed is low, and FIG. 5 AB shows a state where the rotational speed is high. The symbols are common in FIGS. 4 and 5. In these figures, 29 is the same member as the crank connecting rod 29 in FIG. 3. Furthermore, 31, 32 are weights 33, 34, 35, 36
is the arm of the pantograph, 37 and 38 are the sliding pieces, 39
40 and 41 are guide shafts, 42 and 43 are stoppers provided at both ends of the guide shaft 40, 44 and 45 are coil springs, and 46 are arms fixed to the sliding piece 38;
is the hub, 47 is the rotating shaft of the motor, 48 is the motor,
49 is a pin provided on the arm 39, and 50 and 51 are pins 52 and 5 provided on the sliding pieces 37 and 38, respectively.
3 are pins provided on the weights 31 and 32, respectively; 54
1 and 2 respectively show the stepped portions running on the arms 39.

The arms 33, 34, 35, and 36 of the pantograph are connected at both ends with pins 49, 50, 51, and 52 to form a pantograph. A hub 46 is attached to the shaft 47 of the motor.
Guide shafts 40, 41 are attached to the. The sliding pieces 37 and 38 are slidably mounted on a guide shaft 40. Coil springs 44, 45
are inserted between the side surface of the hub 46 and the end surfaces of the slide pieces 37 and 38, so the slide pieces 37 and 38 are normally pushed outward and the stoppers 42,
The outer end surfaces of the sliding pieces 37 and 38 are in contact with and stopped at 43. The weights 31 and 32 are the guide shaft 4
It is attached so that it can slide on top of 1. Since the sliding pieces 37, 38 and the weights 41, 42 are connected by the pantographs 33, 34, 35, 36 as described above, they move while maintaining a constant relationship with each other. That is, when the weights 31 and 32 move outward on the guide shaft 41, the sliding pieces 37 and 38 move inward on the guide shaft 40. In the state shown in Fig. 4, the sliding piece 3 is
7 and 38 are pressed outward by coil springs 44 and 45 and are stopped against stoppers 42 and 43. In that state, the center of the pin 49 is the motor 48.
The center of rotation axis 47 coincides with the center of rotation axis 47 of . When the motor is stopped, the state shown in FIG. 4A is maintained, and in this case, the length of the crank arm is zero. The starting force when starting the motor 48 in this state is found by equation (14), but since the length of the crank arm corresponding to x in equation (14) is zero, the torque when starting the motor 48 is No longer needed.

When the motor 48 starts rotating, centrifugal force acts on the weights 31 and 32, and a force that pulls the weights 31 and 32 outward acts. However, while the rotational speed is low, the force of the coil springs 44 and 45 is greater than the centrifugal force, so the fourth
The state shown in Figure A is maintained. Therefore, motor 4
8's torque is hardly needed. As the rotational speed becomes higher, the centrifugal force acting on the weights 31 and 32 overcomes the force of the coil springs 44 and 45, and the weights 31 and 32 move outward. When the rotational speed becomes higher, the centrifugal force acting on the weights 31, 32 completely overcomes the force of the coil springs 44, 45, and the weights 31, 32 are moved as shown in FIGS. 5A and 5B.
, the sliding pieces 37, 38 move inward.
As shown in FIG. 5B, the stepped portion 54 is in contact with the side surface of the hub 46 and stops without being able to move any further. At this time, the length of the crank arm becomes L as shown in FIG. 5A, and the hammer bank 21 can be oscillated with a predetermined amplitude.

Although no driving force is required to swing the hammer bank 21 at the frequency shown by equation (10),
Driving force is required on the way to reach that rotational speed. The driving force is expressed by equation (9), and this value gradually decreases as ω increases and becomes zero at the frequency of equation (10). Therefore, if the centrifugal force acting on the weights 31 and 32 overcomes the force of the coil springs 44 and 45 at low speeds, a large driving force will be required from the motor 48, which is not a good idea. That is, it is desirable to design such that the centrifugal force acting on the weights 31 and 32 overcomes the force of the coil springs 44 and 45 at a rotation speed as close as possible to the rotation speed expressed by equation (10).

At the frequency of equation (10), the force for swinging the hammer bank 21 is zero, so the crank connecting rod 2
No force is acting on 9. Therefore, at this frequency, the length a of the crank arm shown in FIG. 5 is maintained stably.

However, in a frequency range lower than the frequency of equation (10), the above-mentioned crank arm length a may not be stably maintained. At frequencies lower than equation (10), the driving force is not zero, and a load force is acting on the connecting rod 29. Therefore, when the positional relationship as shown in FIG. 6 is reached, the sliding piece 38 tends to be returned outward by the load force _l acting on the connecting rod 29. Weight 31,3
Since the centrifugal force acting on 2 is not sufficiently large in a region where the rotational speed is low, the sliding piece 38 is returned to the outside. On the other hand, in the positional relationship as shown in FIG. 5A, the load force _l acts in a direction perpendicular to the direction of movement of the sliding piece 38, so that the sliding piece 38 is not returned to the outside.

In such a low rotation speed range, a phenomenon occurs in which the sliding piece 38 swings as the motor 48 rotates. However, as the rotational speed approaches the value of equation (10), the load force _l approaches zero and the centrifugal force also increases, so that the sliding piece 38 no longer swings. In addition,
Even if such swinging of the sliding piece 38 occurs, the sliding piece 3
7. Since the weights 31 and 32 are connected to each other by a pantograph, no unbalanced force is generated and no vibration force is exerted on the outside, and it only occurs for a period of time while the rotational speed increases. This phenomenon is not a problem in practical use.

The hammer bank swinging mechanism of the present invention has been described in detail above, but the requirements for realizing the present invention are summarized as follows. The hammer bank 21 and the balance weight 22 are connected by a spring 23, and the hammer bank 21 and the balance weight 22 are connected by a link 25. The balance weight 22 is attached to the base via a spring 24. Here, the weight of Hammerbank 21
w ₁ , weight of balance weight 22 w ₂ , link 25
When the arm lengths of are l ₁ and l ₂ (see Figure 2),
Each is determined so that the relationship w ₁ /w ₂ =l ₂ /l ₁ holds true. Furthermore, the required swing frequency of the hammer bank 21 is determined by ω, and the spring constant of the spring 23 is k ₁ Spring 2
When the spring constant of 4 is k ₂ and g is the acceleration of gravity, ω ² = {k ₁・(1+l ₂ /l ₁ ) ² +k ₂・(−l ₂ /l ₁ ) ² }/{w ₁ / g・(1+l
₂ /l ₁ )} are determined so that the following holds true. The hammer bank 21 is connected to the crank mechanism 3 by a connecting rod 29.
Connected to 0. In the crank mechanism 30, the length of the crank arm is zero when the rotation speed is low;
When the rotation speed approaches the required oscillation frequency, the weight 3
It is constructed of a mechanism in which the length of the crank arm is adjusted to the required length by the monkey force acting on the arms 1 and 32.

In the hammer bank swinging mechanism with such a configuration, no vibration force acts on the base at any swing frequency.

No driving force is required for rocking at the required rocking frequency.

The torque required for the motor at startup can be extremely small.

This provides the following features.

Next, an actual concrete example will be explained.

Now consider the case where the hammer bank 21, which weighs 800 g _r , is oscillated in a sinusoidal motion at 40 Hz. As mentioned above, increasing the weight of the balance weight 22 will increase the weight of the device, and conversely, decreasing the weight will increase the swing amplitude of the balance weight 22, which may not be a good idea in terms of device configuration. do not have. Therefore, here, the weight of the balance weight 22 is calculated as the weight of the hammer bank 2.
Let's set it to 800g _r , which is equal to the weight of 1.

In order to prevent the vibrations caused by the swinging of the hammer bank 21 from being transmitted to the base, the lengths l ₁ and l ₂ of the arms of the link 25 must be determined according to equation (8). In this example, l ₁ = l ₂ must be satisfied.

Next, the spring constants of springs 23 and 24 must be selected. There are an infinite number of combinations to choose from, but in order for the springs 23 and 24 to have the role of holding the hammer bank 21 and the balance weight 22 at the same time as the springs 23 and 24, it is important not to make either spring constant too small. Undesirable. Therefore, we will select k ₁ = k ₂ here. l ₁ = l ₂ , k ₁ determined from the above selections
Substituting the relationship =k ₂ into equation (10) yields ω ² =5g/2·k ₁ /w ₁ (15). In this formula, ω=2π×40rad/sec, w ₁ =
Substituting 0.8Kg g=980cm/sec ² to find k ₁ gives k ₁ = 20.6Kg/cm .

Therefore, the spring constant of springs 23 and 24 is 20.6.
By setting the value to Kg/cm, driving force can be made unnecessary at an oscillation frequency of 40Hz. However, since two springs 23 and 24 are used each,
The spring constant per sheet must be 10.3Kg/cm.

Note that the arm lengths l ₁ and l ₂ of the link 25 can be arbitrarily selected according to the conditions of the mechanism as long as l ₁ =l ₂ is satisfied.

Next, the crank mechanism 30 will be explained. In order to swing the hammer bank 21 at 40Hz, it is necessary to rotate the crank mechanism 30 at 40r, p, and s. At this time, the centrifugal force acting on the weights 31 and 32 must be strong enough to hold the crank arm length at a predetermined value so as not to fluctuate due to disturbances. be. now,
It is assumed that the design is such that the force _l pressing the sliding piece 38 inward is 3 kg at rotational speeds of 40 r, p, and s. In addition, the centrifugal force causes the coil springs 44, 4
It was previously stated that it is desirable that the rotational speed at which the force of 5 is overcome is as close as possible to the required oscillation frequency, but here the rotational speed ωb is set to 30Hz.

Now, the total weight of the weights 31 and 32 is w, the return force of the coil springs 44 and 45 is k, respectively, and the weight of the weight 3 is
If the distance from the center of gravity of the weights 1 and 32 from the center of the rotating shaft 47 of the motor 48 is r, then the centrifugal force cb acting on each weight 31 or 32 at the rotation speed ωb is cb=w/g・r・ωb ² ...(16) becomes. w and r so that this value is equal to the return force k per coil spring 44 or 45.
decide.

Further, the centrifugal force c at the required rocking frequency ω is c=w/g·r·ω ² (17). w and r are determined so that this value is equal to the return force k per coil spring 44 or 45 plus the aforementioned l.

That is, each value is determined so as to satisfy the following simultaneous equations: w/g・r・ωb ² =k (18) w/g・r・ω ² =k+l (19).
As a result, w・r=g/ω ² −ωb ²・l (20) k=ωb ² /ω ² −ωb ²・l (21). In equation (20), the above-mentioned value ω=2π×40rad/
By substituting secωb=2π×30rab/sec l=3Kg, we get w・r=105g ^r −cm.

Now move the weight 31 or 32 from the center of the motor 48.
If the distance r to the center of gravity of is 2 cm, then w is 52.5 g.
It becomes ^r . Therefore, the weight per weight is 26.25g ^r
And it is sufficient. Now these weights 31 and 32 have a specific gravity of 8.9.
If it were made of copper, it would be cylindrical with a diameter of 1.6 cm and a thickness of 1.5 cm. Also, from equation (21), k=3.8Kg. That is, it is sufficient to provide coil springs 44 and 45 that each generate a force of 1.9 kg.

As mentioned above, the driving force m for swinging the Heimerbank 21 is given by equation (9), but when a crank mechanism like the one of the present invention is not provided, m=
A starting force of 5・k ₁・x is required, but if a mechanism is provided in which the length of the crank arm is maintained at zero up to a certain frequency as in the present invention, m=5・x.
k ₁・x−2・w/g・ω ²・x is required.

Applying the results of the above calculations to these formulas and finding their upper value, we get m'/m=12〓/5k ₁・ω ² =0.44, and as in the calculation above, the length of the crank arm up to 30Hz. It can be seen that if a is kept at zero, the required driving force can be reduced to 1/2 or less.

Note that if the frequency at which the length a of the crank arm is maintained at zero is set higher, the starting force can be further reduced.

As explained above, in the hammer bank swing mechanism of the present invention, the vibration force accompanying the swing of the hammer bank is perfectly balanced, and at the same time, the force required for driving can be reduced. It can make printers more practical.

In the above description, flat springs 23 and 24 are shown, but it is clear that any shape can be used as long as the springs have a predetermined spring constant. In particular, it is advantageous to adopt a bent shape as shown in FIG. 7 because the overall size can be reduced.

Note that the springs 23 and 24 do not need to be plate springs, and for example, coil springs may be used. However, in the case of a coil spring, since it does not have a function of supporting weight, it is necessary to separately provide a slide mechanism for supporting the hammer bank 21 and the balance weight 22.

Furthermore, in the above explanation, we have explained the case where the length of the crank arm is maintained at zero when the rotation speed is below a predetermined value, but as is clear from equation (9), if x is small, the driving force m is relatively small. is reduced, so even if the length of the crank arm is not zero when the rotational speed is low, as long as it is shorter than a predetermined length, the effect of reducing the driving force can be obtained.

[Brief explanation of the drawing]

Fig. 1 is a diagram explaining the printing method of a dot matrix type line printer, Fig. 2 is a diagram showing the principle of the balance mechanism of the hammer bank swing mechanism of the present invention, and Fig. 3 is a diagram illustrating an embodiment of the present invention. 4A to 5C, 5A, B, and 6 are diagrams showing one embodiment of a crank mechanism for swinging the hammer bank and its operation, and FIG. 7 is a diagram illustrating another embodiment of the invention. It is a perspective view showing an example of. 21 is a hammer bank, 22 is a balance weight, 23 and 24 are leaf springs, 25 is a link, 29 is a crank connecting rod, 30 is a crank mechanism, 31,
32 is a weight; 33, 34, 35, and 36 are arms of the hunter graph; 40, 41 are guide shafts; 44, 45 are springs; 47 is a rotating shaft of a motor; and 48 is a motor.

Claims (1)

[Claims] 1. In a mechanism for swinging a hammer bank of a dot-trick type line printer at a constant frequency, a balance weight is provided, and the balance weight and the hammer bank are connected by a first spring. The balance weight is connected to the base by a second spring, and the hammer bank and the balance weight are connected by a link that can rotate around a fixed rotation center. and the weight of the hammer bank w ₁ The weight of the balance weight is w ₂ The length of the arm from the center of rotation to the hammer bank side in the above link is l ₁ The length of the arm from the center of rotation to the balance weight side When the length is l ₂ , the first spring constant is K ₁ , the spring constant of the second spring is K ₂ , and the swing frequency of the hammer bank is ω, W ₁ /W ₂ = l ₂ /l _1. ω ² = {K ₁ (1+l ₂ /l ₁ ) +k ₂ (l ₂ /l ₁ ) ² }/{ω ₁ /g(1+l ₂ /l ₁
)} Each constant is determined to satisfy the following two equations, and furthermore, the hammer bank is connected to a crank mechanism, and when the rotation speed is below a predetermined value, the crank arm length is However, when the number of rotations exceeds a predetermined value, a mechanism is provided in which the length of the crank arm is maintained at a predetermined value by centrifugal force acting on the weight. A hammer bank rocking mechanism characterized by: