JPH11194006A - Angle measuring apparatus - Google Patents

Abstract

PROBLEM TO BE SOLVED: To obtain an angle measuring apparatus in which it is possible to prevent a rotation angle detected in the boundary between regions for matching from becoming discontinuous, by a method wherein a region for matching is installed between voltage regions of a potentiometer, a weighting factor according to the distance between both adjacent regions is multiplied in a point inside the region for matching, and he rotation angle is detected. SOLUTION: A rotation angle θ1 corresponding to the output voltage V2 of a second potentiometer is detected in the voltage region A1 of a twin potentiometer, A rotation angle θ2 corresponding to an output voltage V1 is detected in a region A2 . A rotation angle θ2 , in which 360 deg. is subtracted from the rotation angle θ2 is detected in a region A2 '. Then, a region A3 for matching is installed between the region A1 and the region A2 '. A weighting factor is found in a point P inside the region A3 according to the ratio of electric power widths with reference to the electric power width -Va -L1 between a point Q5 and a point Q2 . The angles θ2 ', θ1 which are found by supposing that the point P is situated in the region A2 ' or the region A1 are multiplied respectively by weighting factros. The sum of both is regarded as a rotation angle θ3 in the region A3 for matching. By the same method, a rotation angle θ4 in a region A4 for matching is detected.

Description

DETAILED DESCRIPTION OF THE INVENTION

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to an angle measuring device using a 360 ° detecting dual potentiometer, and more particularly to dead zone processing.

[0002]

2. Description of the Related Art As a conventional technique, an angle measuring apparatus which was considered before the present invention was obtained will be described. 360 ° detection dual potentiometer 5 used for this device
As shown in FIG. 6, potentiometers (hereinafter referred to as potentiometers) 1 and 2 are arranged so as to be shifted from each other by 180 °.
When the common rotating shaft 3 provided at the axis of each of the ring-shaped resistors 1b and 2b is rotated by θ by an external force,
The respective sliders 1b and 2b work in conjunction with each other to
a, 1b, and outputs a voltage obtained by dividing the power supply voltage −V ₀ to + V _{0 according} to the rotation angle θ.

The power supply voltages + V ₀ , −V ₀ are applied to both ends of the resistors 1a, 1b of each potentio. Resistors 1a, 1
between both ends of b, without resistor, angle width 10 °
A small area is a dead zone. When one potentiometer enters the dead zone, the output value of the other potentiometer can be used to detect an angle of 0 ° to 360 °. Output voltages V ₁ and V _{2 of} potentiometers 1 and ₂ and sliders ₁ b and 2
FIG. 8 shows the relationship between b (rotation axis 3) and the rotation angle θ. The characteristics of potentio 1 are indicated by broken line ABCD, and the characteristics of potentio 2 are indicated by broken line AEFGD. α is potentiometer 1
Α ₁ + α ₂ is the dead zone width of the potentiometer 2, and α ₁ = α ₂ = α / 2.

When the output voltage of potentiometer 1 is V₁= V_1a(Q
₁), The output voltage of potentiometer 2 is V_Two= V
_2a(Q_ThreePoint) and the output voltage of potentiometer 2 is V_Two= V
_2b(Q_Two), The output voltage of potentiometer 1 is V₁=
V_1b(Q_FourPoint). V_2a≤V _Two<V_2bV_1b≤V₁
≦ 0; 0 ≦ V₁<V_1aRange of each
Area A₁, A_Two, A_Two'.

The rate of change of the angle θ with respect to the voltage on the straight line EG is (360−α) / 2V ₀ . Straight lines AB and CD
The rate of change of θ above is the same value. In the area A ₁ , when the coordinates of an arbitrary point H are (θ ₁ , V ₂ ) and the length of the base of the triangle EHI is represented by EI, the detection angle θ ₁ corresponding to V ₂ is given by the following equation (1). More required.

[0006] _{θ 1 = α 1 + EI =} α 1 + {(360-α) / 2V 0} (V 0 + V 2) ... (1) As shown in FIG. 9, in the region A _2, any point Pa The coordinates are (θ ₂ , V ₁ ). Set the straight line AB to α of the dead zone width.
Let A 'and B' be straight lines which are shifted laterally in parallel only, point J which is shifted point Pa upward by 2V ₀ is on an extension of A'B '. If the length of the base of the triangle A′JK is represented by A′K, the detection angle θ ₂ corresponding to V ₁ can be obtained by the following equation (2).

Θ ₂ = α + A′K = α + ｛(360−α) / 2V ₀ ｝ (V ₁ + 2V ₀ ) (2) As shown in FIG. 9, in the area A ₂ ′, an arbitrary point on the straight line AB The coordinates of the point Pb are (θ ₂ ′, V ₁ ). The Pc point obtained by translating the Pb point by 360 ° in the horizontal direction is on an extension of the straight line CD. If the coordinates of the Pc point are (θ ₂ , V ₁ ), θ ₂ = θ ₂ ′ + 360 ° (3) Instead of directly finding θ ₂ ′, first find θ ₂ , then θ ₂ > 36
At _{0 °, θ 2 '= θ} 2 -360 from _{° ... (4), θ 2} ' Request.

The angle θ ₂ of the Pa point in the above-mentioned A ₂ region
Similarly, when the point L (θ ₂ , V ₁ + 2V ₀ ) is obtained by shifting the Pc point upward by 2V ₀ , and the length of the base of the triangle A′LM is represented by A′M, The detection angle θ ₂ corresponding to the point V ₁ is obtained by the following equation (5). θ ₂ = α + A′M = α + ｛(360−α) / 2V ₀ ｝ (V ₁ + 2V ₀ ) (5) Equation (5) is exactly the same as equation (2), which is convenient for calculation.

In the angle measuring device 10, as shown in FIG.
Output voltage V of tension 1, 2₁, V _TwoA / D converter
Digital data DV with the data 7₁, DV_TwoTo each
Input to the angle calculation unit 9 and the detected angle θ (θ₁, Θ
_Two, Θ_Two').

[0010]

Consider a detection angle θ at the boundary between the areas A ₁ and A ₂ ′. _First , θ ₁ is obtained from the equation (1) applied to the area A ₁ . In equation (1), V ₀ =
Substituting 2.5 V, V ₂ = V _2a = −1.0 V, θ ₁ = α ₁ + {(360−α) / 2V ₀ } (V ₀ + V _2a ) = α ₁ + 108−0.3α = 108 + 0 .2α (α ₁ = α / 2) = 110 (degrees) (assuming α = 10) (6) Next, θ ₂ is obtained from the equation (2) applied to the area A ₂ .
Substituting V ₁ = V _1a = 1.55V into the equation (2), θ ₂ = α + {(360−α) / 2V ₀ } (V _1a + 2V ₀ ) = α + 471.6-1.31α = 471.6 −0.31α = 468.5 (degrees) (α = 10) Since θ ₂ > 360 (degrees), θ ₂ ′ is obtained from equation (4).

Θ ₂ ′ = θ ₂ -360 = 108.5 (degrees) (7) From (6) and (7), it can be seen that there is a slight difference between θ ₂ ′ and θ _1. . The detection angle at the boundary of the area is
Ideally, using equation (1) or equation (3),
The same result must be obtained. However, since the potentiometers 1 and 2 have manufacturing variations, the output voltage values V ₁ and V ₂ involve errors. Therefore, as in the example described above, the detection angle θ is discontinuous at the boundary between A ₁ and A ₂ ′. The same applies to the boundary of the area A ₁ and A _2.
When the control system is controlled by the discontinuous detection angle θ, the control becomes unstable. An object of the present invention is to perform processing so that a discontinuity does not occur in a detection angle at a boundary between regions.

[0012]

According to a first aspect of the present invention, there is provided a dual potentiometer having a rotating shaft rotated by an external force, and first and second divided voltages output from the dual potentiometer. The present invention relates to an angle measuring device that calculates the rotation angle of the rotating shaft from V ₁ and V ₂ . The dual potentiometer is such that the first and second potentiometers are arranged 180 ° apart from each other about a common axis of rotation, and the first and second potentiometers
Positive and negative power supply voltages + V ₀ and −V ₀ are applied to both ends of the second potentiometer across a dead zone (α degrees).

When the divided voltage V ₁ of the first potentiometer is V
_The rotation angle θ at which ₁ = 0 is set to θ = 0. When the rotating shaft makes one rotation (360 ° counterclockwise from θ = 0 °), the first potentiometer changes the divided voltage V ₁ linearly from 0 to V ₀ and then passes through the dead zone (α). It varies linearly from -V ₀ 0 Te. When the rotating shaft makes one rotation, the second potentiometer changes the divided voltage V ₂ to the dead zone (α /
After changing linearly in + V ₀ from -V ₀ through 2),
It becomes a dead zone (α / 2).

V₀Less positive voltage V_aAnd V
₁= V_aV when_TwoThe measured value of L ₁And V_Two= V_a
V when₁The measured value of L_TwoAnd -V_a≤V_Two≤V
_a(Area A₁), Θ₁= Α / 2 + ｛(360−α) / 2V₀｝ (V₀+ V
_Two) (Degree) to calculate the rotation angle. -V_a≤V₁≤0 range (territory
Area A_Two), Θ_Two= Α + ｛(360−α) / 2V₀｝ (V₁+2
V₀) (Degrees) to calculate the rotation angle. 0 ≦ V₁≤V_aRange (area
A_Two'), Θ_Two'= Θ_TwoThe rotation angle is calculated from -360 (degrees). L₁≤V_Two≤-V_aRange
(Area A_Three), Γ₁= (V_Two-L₁) / (− V_a−
L₁) And θ_Three= Γ₁θ₁+ (1-γ₁) Θ_Two′ Calculate with rotation angle. L_Two≤V₁≤-V_aRange
(Area A_Four), Γ_Two= (V₁-L_Two) / (− V_a−
L_Two), Θ_Four= (1-γ_Two) Θ₁+ Γ_Twoθ_Two The rotation angle is calculated more.

[0015]

DESCRIPTION OF THE PREFERRED EMBODIMENTS In the present invention, as shown in FIG.
In the area A₁And A_Two'And A₁And A_TwoBetween
Area A for joint use_ThreeAnd A_FourAnd the detection angle
By devising the calculation formula of degree, the detection angle θ is not
I try not to make a continuation. Potential 1 θ-V characteristics
V in sex₁= V_a(= 1.05V) ₁
And the output voltage of potentiometer 1 is V₁= V_aWhen
Output voltage of tension 2 is V_Two= L₁(Point Q_FiveVoltage value)
And On the θ-V characteristic of potentiometer 2, V_Two=
-V_a(= -1.05V)_TwoAnd Pote
On the θ-V characteristic of the_Two= V_a(= 1.
05V)_ThreeAnd the output voltage of potentiometer 2
Is V_Two= V_aWhen the output voltage of the potentiometer 1 is L
_Two(Point Q₆Voltage value). Potential 1 θ-V characteristics
Sexually, V₁= -V_a(= 1.05V)
To Q_FourAnd Α, α₁, Α_TwoIs the dead zone width
And α₁= Α_Two= Α / 2.

-V_a≤V_Two≤V_a ... (8)₁And −V_a≤V₁≦ 0 (9)_TwoAnd 0 ≦ V₁≤V_a ... the voltage range of (10)_Two'And L₁≤V_Two≤-V_a ... the voltage range of (11)_ThreeAnd L_Two≤V₁≤-V_a ... the voltage range of (12)_FourAnd (1) Area A₁At
Is the output voltage V_TwoDetection angle θ corresponding to₁Is the straight line in FIG.
The point H on the EG (θ₁, V_Two). θ₁Is
Conventional area A₁In the same manner as in the above,
Desired. If rewritten, θ₁= Α₁+ EI = α₁+ ｛(360-α) / 2V₀｝ (V₀+ V_Two) (Degrees) (1) (2) Area A_Two, The output voltage V₁Inspection corresponding to
Outgoing angle θ₁To the point Pa (θ on the straight line CD in FIG._Two, V₁)
It is represented by. θ_TwoIs the area A of the conventional example_TwoSame as
By the way, it is obtained by the above-mentioned equation (2). In other words, θ_Two= Α + A′K = α + ｛(360−α) / 2V₀｝ (V₁+ 2V₀) (2) (3) Area A_Two', V_TwoAnd θ₁Figure 4
Pb (θ on the straight line AB_Two', V₁)
You. Conventional area A_Two', The point Pb
The point Pc (θ_Two, V₁), Point Pc further up
The moved point L (θ _Two, V₁+ 2V₀), Θ_Two= Θ
_Two'+ 360 ° is the same as the previously described equation (2).
Request. If rewritten, θ_Two= Α + A′M = α + ｛(360−α) / 2V₀｝ (V₁+ 2V₀)… (5) θ_Two> 360, so θ_Two'= Θ_Two-360 (degrees) (4) (4) Area A_Three, The straight line Q in FIG.
_FiveQ_TwoThe upper point P (Q_Three, V_Two) Detection angle θ_ThreeAsk for
Shall be. Voltage width V between point P and point Q5_Two-L₁Is point Q
_FiveAnd point Q _TwoVoltage width -V_a-L₁Ratio γ₁= (V_Two-L₁) / (− V_a-L₁) ... (13) is, for example, γ₁= 0.1 (10%), the point P is
Area A₁Far from area A _Two’Is very close to
And the point P is in the area A_Two'(5), (4)
Angle θ obtained from the formula_Two'To 1-γ₁= Weight 0.9
The beam angle (1-γ ₁) Θ_Two'And point P is in area A₁It is in
And the angle θ obtained from equation (1)₁To γ₁= 0.1
Weighted angle γθ₁Corresponds to the point P
Detection angle θ_ThreeAnd That is, θ_Three= Γ₁θ₁+ (1-γ₁) Θ_Two'(14) In this way, the point P becomes the area A_Two'And A_ThreeOf the border with
Point Q_FiveWhen γ₁= 0, θ_Three= Θ
_Two'And the area A_TwoMatches the detection angle at the upper limit of ′
I do. Point P is in area A_ThreeAnd A₁Point Q at the boundary with_TwoExist
When present, γ ₁= 1, so θ_Three= Θ₁Becomes
Area A₁Coincides with the detection angle at the lower limit of. like this
In addition, there is no discontinuity in the detection angle at both boundaries. (5)
Area A_FourAlso in the area A_ThreeIn the same way as in FIG.
Point P (θ_Four, V_Two). Points P and Q₆Electricity between
Pressure range (V₁-L_TwoPoint Q)₆, Q_FourVoltage width between -V_a
-L_TwoRatio γ_Two γ_Two= (V₁-L_Two) / (− V_a-L_Two)... (15), the point P is set to the area A₁Θ assuming that₁
Weight (1-γ_Two), And the point P is in the area A_TwoIt is temporary
Θ_TwoWeight γ_TwoAnd θ_Four= (1-γ_Two) Θ₁+ Γ_Twoθ_Two ... (16) is the detection angle.

If the point P is located at the lower limit Q ₆ of the area A ₄ , γ ₂ = 0 and θ ₄ = θ ₁ .
Matches the value of ₁ upper limit. If the point P is located at the upper limit Q ₄ of the area A ₄ , γ ₂ = 1, θ ₄ = θ ₂ , and the area A 4
Matches the lower limit of ₂ . Thus, areas A ₁ and A ₄
It does not occur discontinuity of the detection angle of the boundary, and A ₄ and A ₂ of the boundary.

[0018]

According to the present invention, the voltage range A of the potential
A region A ₃ for alignment is provided between ₂ ′ and A _1, and a region A ₃
Is a weight coefficient (1−γ ₁ ) and γ ₁ according to the proximity to both adjacent areas A ₂ ′ and A _1, and point P is an area A
₂ 'or angle theta ₂ was determined on the assumption that in A _1', theta
₁ is multiplied, and the detection angle θ ₃
And

[0019] Also provided area A ₄ for alignment between the regions A ₁ and A _2, performs the same processing, the discontinuity by these processes, the detection angle at both ends boundary area A _3, A ₄ Can be prevented from occurring.

[Brief description of the drawings]

FIG. 1 is a graph showing output voltage characteristics and a voltage range of a potentiometer with respect to a rotation angle of a dual potentiometer used in the angle measuring device of the present invention.

FIG. 2 is a graph for explaining how to obtain a detection angle θ ₁ with respect to an output voltage V ₂ in an area A ₁ in FIG.

FIG. 3 is a graph for explaining how to obtain a detection angle θ ₂ with respect to an output voltage V ₂ in an area A ₂ in FIG. 1;

FIG. 4 is a graph for explaining how to obtain a detection angle θ ₂ ′ with respect to an output voltage V ₂ in an area A ₂ ′ of FIG.

Figure 5 is a graph illustrating a method of obtaining detection angle theta ₃ and theta ₄ for regions A ₃ and the output voltage V ₂ and V ₁ of the the A ₄ in FIG. 1.

FIG. 6 is a diagram showing a basic configuration of a double potentiometer.

FIG. 7 is a block diagram of an angle measuring device according to the related art and the present invention.

FIG. 8 is a graph showing output voltage characteristics and a voltage range of a potentiometer with respect to a rotation angle of a dual potentiometer used in a conventional angle measuring device.

9 is a graph for explaining how to obtain detection angles θ ₂ and θ ₂ ′ with respect to output voltages V ₁ in regions A ₂ and A ₂ ′ in FIG.

Claims (1)

[Claims] 1. A power supply having a rotary shaft rotated by an external force.
Having a continuous potentiometer, its dual potentiometer
Output first and second divided voltages V₁, V_Twobefore that
In the angle measuring device for calculating the rotation angle of the rotating shaft, the double potentiometer may include a first potentiometer and a second potentiometer.
Meters are 180 ° shifted from each other about a common axis of rotation
Dead zones of the first and second potentiometers
(Α degrees) between the positive and negative power supply voltage + V₀And -V
₀Is applied, and the divided voltage V of the first potentiometer is applied.₁Is V₁= 0 and
Is set to θ = 0, and the first potentiometer rotates the rotation axis once (θ = 0).
(360 ° counter-clockwise), the divided voltage V
₁Is 0 to V₀After changing linearly, the dead zone (α)
Through -V₀From 0 to 0, and the rotation axis of the second potentiometer makes one rotation.
And the divided voltage V_TwoIs -V through the dead zone (α / 2)₀Or
+ V₀After changing linearly to the dead zone (α / 2)
And V₀Less positive voltage V_aAnd V₁= V_aNoto
Mushroom V_TwoThe measured value of L ₁And V_Two= V_aV when₁
The measured value of L_TwoAnd −V_a≤V_Two≤V_a(Area A₁)₁= Α / 2 + ｛(360−α) / 2V₀｝ (V₀+ V
_Two) (Degrees) to calculate the rotation angle,_a≤V₁≤0 (region A_Two)_Two= Α + ｛(360−α) / 2V₀｝ (V₁+2
V₀) (Degrees) to calculate the rotation angle, 0 ≦ V₁≤V_a(Area A_Two′)_Two'= Θ_TwoCalculate the rotation angle from -360 (degrees)₁≤V_Two≤-V_a(Area A_Three) Is γ₁= (V
_Two-L₁) / (− V_a-L₁) And θ_Three= Γ₁θ₁+ (1-γ₁) Θ_Two′ To calculate the rotation angle, L_Two≤V₁≤-V_a(Area A_Four) Is γ_Two= (V
₁-L_Two) / (− V_a-L_Two), Θ_Four= (1-γ_Two) Θ₁+ Γ_Twoθ_Two Angle measuring device characterized by calculating the rotation angle more
Place.