CN101196427A - Method for calibrating and measuring unbalance amount of tire - Google Patents

Method for calibrating and measuring unbalance amount of tire Download PDF

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CN101196427A
CN101196427A CNA200610129929XA CN200610129929A CN101196427A CN 101196427 A CN101196427 A CN 101196427A CN A200610129929X A CNA200610129929X A CN A200610129929XA CN 200610129929 A CN200610129929 A CN 200610129929A CN 101196427 A CN101196427 A CN 101196427A
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mrow
mover
unbalance
tire
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CN100582710C (en
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张芝泉
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Abstract

The invention discloses a tyre unbalance amount marking and measuring method, which comprises a tyre unbalance amount marking part being composed of amount marking and eccentricity revising and a tyre unbalance amount measuring part. Wherein, the work finished in the amount marking part is to count the proportion factor between the voltage value measured by the piezoelectric sensor and the practical force worked on the tyre; the work finished in the eccentricity revising part is to use the proportion factor obtained in amount marking to measure and count the unbalance amount of a main shaft and a rim system; the measuring of the tyre unbalance amount is that: according to the proportion factor measured in amount marking and the unbalance amount of the main shaft and the rim system, the practical unbalance amount of tyre is measured and calculated. The invention improves the reliability and stability of data sampling and also can conduct rectifying a deviation automatically, therefore improving the measuring precision and repeating precision of the tyre unbalance amount and reducing error.

Description

Tire unbalance calibration and measurement method
Technical Field
The invention relates to a method for measuring the unbalance amount of a tire. In particular to a tire unbalance calibration and measurement method which improves the reliability and stability of data sampling and can automatically correct the rim, thereby improving the measurement precision of the tire unbalance and reducing errors.
Background
The dynamic balance of the tire refers to different degrees of centrifugal force in each direction when the tire rotates, and is an important parameter for checking the quality of the tire. When the dynamic balance state of the tire is good, the stress in each direction is basically the same in the rotation process of the tire, and when the dynamic balance state is not good, the centrifugal force in a certain direction is too large or too small, so that the quality of the tire is influenced, and when the tire is seriously installed on an automobile, the tire is easily blown out during operation, and safety accidents are caused. Therefore, the general tire needs to be subjected to a dynamic balance test before leaving the factory or after maintenance, so that the quality of the tire is ensured, and potential safety hazards are eliminated. Theoretically modeling and analyzing the dynamic balancing machine relies on the following assumptions: (1) maintaining a strict linear relationship between the voltage and the force of the piezoelectric sensor; (2) the amplitude and the phase of the voltage are measured absolutely accurately; (3) the dynamic balancing machine is absolutely accurate to manufacture and assemble and does not change in the rotating process; (4) the balance rotating speed is relatively stable. For various reasons, these assumptions are difficult to guarantee absolutely in practice, and there are errors in the measurement. Although these errors can be theoretically analyzed, they are very complicated and difficult to measure, so the dynamic balancing machine must be calibrated. Because the main shaft of the dynamic balancing machine is connected with the driving motor through the belt and the belt pulley, and the lower piezoelectric sensor is close to the belt, the vibration of the belt can cause certain influence on the acquisition of lower channel signals, and although most of interference is reduced through means of filtering of software, hardware and the like, the influence still can be generated on measurement data, which shows that the lower unbalance of the tire is not as high as the measurement accuracy of the upper unbalance.
Disclosure of Invention
The invention aims to solve the technical problem of providing a tire unbalance calibration and measurement method which improves the reliability and stability of data sampling and can automatically correct the rim, thereby improving the measurement accuracy of the tire unbalance and reducing errors.
The technical scheme adopted by the invention is as follows: a tire unbalance calibration and measurement method comprises two parts of tire unbalance calibration and tire unbalance measurement, wherein the two parts are formed by quantity calibration and eccentricity correction, and the method comprises the following steps: what is done in the calibration of the quantities is to calculate a proportionality coefficient between the voltage value measured by the piezoelectric sensor provided on the spindle and the force actually acting on the tire; what is accomplished in the eccentricity correction is that the inherent unbalance of the main shaft and the rim system is measured and calculated by using the proportionality coefficient calculated in the amount calibration; the unbalance of the tire is measured and calculated according to the proportionality coefficient measured in the quantity calibration and the unbalance of the main shaft and the rim system obtained in the eccentricity correction.
The quantity calibration comprises the following steps: assigning values to the form parameters according to the measurement results; calculating coefficients from the assigned values
Figure A20061012992900061
(ii) a According to the coefficient
Figure A20061012992900062
Calculating coefficients <math><mrow> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mo>=</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>11</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>22</mn> </msub> <mo>-</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>12</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>21</mn> </msub> <mo>;</mo> </mrow></math> According to the coefficient
Figure A20061012992900064
Calculating the proportionality coefficient
The form parameter assignment according to the measurement result comprises that when no weight is added to the tire, the following voltage values on the tire are measured respectively: uu0, Φ u0, Ud0, Φ d 0; when weights are added on the upper surface of the tire, the voltage values of the upper surface and the lower surface of the tire are respectively measured as follows: uu1, Φ u1, Ud1, Φ d 1; when weights are added below the tire, the voltage values above and below the tire are respectively measured: uu2, Φ u2 and Ud2, Φ d 2.
Calculating coefficients according to the assigned values
Figure A20061012992900066
The method comprises the following steps: calculate out <math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>11</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>1</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo></mo> </mrow> <mo>/</mo> <mn>100</mn> <mo>=</mo> <msub> <mrow> <mi>K</mi> <mn>11</mn> </mrow> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>1</mn> </mrow></math> A stage (2); calculate out <math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>12</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>2</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo></mo> </mrow> <mo>/</mo> <mn>100</mn> <mo>=</mo> <mi>K</mi> <mn>1</mn> <msub> <mn>2</mn> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>2</mn> </mrow></math> A stage (2); calculate out <math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>21</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>1</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo></mo> </mrow> <mo>/</mo> <mn>100</mn> <mo>=</mo> <msub> <mrow> <mi>K</mi> <mn>21</mn> </mrow> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>3</mn> </mrow></math> A stage (2); calculate out <math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>22</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>2</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo></mo> </mrow> <mo>/</mo> <mn>100</mn> <mo>=</mo> <msub> <mrow> <mi>K</mi> <mn>22</mn> </mrow> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>4</mn> </mrow></math> And (3) a stage of (a).
The coefficient of dependence
Figure A200610129929000613
Calculating coefficients
Figure A200610129929000614
The method comprises the following steps: <math><mrow> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mo>=</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>11</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>22</mn> </msub> <mo>-</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>12</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>21</mn> </msub> <mo>.</mo> </mrow></math>
the coefficient of dependence
Figure A200610129929000616
Calculating the proportionality coefficient
Figure A200610129929000617
The method comprises the following steps:
Figure A200610129929000618
Figure A200610129929000619
Figure A200610129929000621
the eccentricity correction comprises the following steps: measuring voltage values of 8 groups of upper and lower piezoelectric sensors in the process that the tire rotates for one circle relative to the rim, summing 8 groups of measurement data vectors, and averaging to obtain a voltage value caused by the imbalance of the main shaft and the rim; and converting the voltage value into a corresponding force phase.
In the stage of converting the voltage value into the corresponding force, firstly, the voltage value corresponding to the unbalance between the upper plane and the lower plane of the rim system, which is measured by the piezoelectric sensor, is set; setting the unbalance amount of the upper plane and the lower plane of the main shaft and the rim system at the position of the correction weight; the corresponding forces are thus calculated as:
ML″=K1×Uuw+K2×Udw
MR″=K3×Uuw+K4×Udw
the measurement of the unbalance amount of the tire comprises the following steps: converting the voltage value measured by the piezoelectric sensor into corresponding force; calculating the unbalance of the tire at the position of a weight according to the unbalance of the spindle and the rim system obtained by eccentric correction; converting the unbalance of the tire at the position of the weight into the unbalance at the position of the seam allowance, namely a final unbalance value; and calculating the static unbalance and the even unbalance according to the upper unbalance and the lower unbalance.
The step of calculating the unbalance amount of the tire at the position of the weight according to the unbalance amount of the spindle and the rim system obtained by eccentric correction is that firstly, the unbalance amounts of the upper plane and the lower plane of the spindle and the rim system at the position of the weight are respectively set as: mL″,βL"and MR″,βR"; and then the measured unbalance amounts of the upper plane and the lower plane of the tire at the position of the weight are respectively set as: mL,βLAnd MR,βR(ii) a Will ML′,βL' and MR′,βR' and ML″,βL"and MR″,βR"the respective vectors are subtracted to obtain the upper and lower unbalance of the tire itself:
unbalance on tire: mL=ML′-ML″
Lower tire unbalance amount: mR=MR′-MR″
The stage of converting the unbalance of the tire at the weight position into the unbalance at the bead position, namely the final unbalance value, comprises calculating the unbalance of the upper surface and the lower surface of the tire bead position as m respectively according to the measuring meterl,θlAnd mr,θrWherein:
<math><mrow> <msub> <mover> <mi>m</mi> <mo>&OverBar;</mo> </mover> <mi>l</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>a</mi> <mo>+</mo> <mi>b</mi> </mrow> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>-</mo> <mfrac> <mi>c</mi> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>)</mo> </mrow> <mfrac> <msub> <mi>R</mi> <mn>1</mn> </msub> <msub> <mi>R</mi> <mn>2</mn> </msub> </mfrac> <mo>,</mo> </mrow></math> <math><mrow> <msub> <mover> <mi>m</mi> <mo>&OverBar;</mo> </mover> <mi>r</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>b</mi> <mo>+</mo> <mi>c</mi> </mrow> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>-</mo> <mfrac> <mi>a</mi> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>)</mo> </mrow> <mfrac> <msub> <mi>R</mi> <mn>1</mn> </msub> <msub> <mi>R</mi> <mn>2</mn> </msub> </mfrac> <mo>.</mo> </mrow></math>
the tire unbalance calibration and measurement method improves the reliability and stability of data sampling, and can automatically correct the rim, thereby improving the measurement precision and the repeatability precision of the tire unbalance and reducing errors.
Drawings
FIG. 1 is a flow chart of quantity calibration;
FIG. 2 is a flow chart of eccentricity correction;
FIG. 3 is a tire unbalance amount calculation flowchart;
fig. 4 is a tire dynamic balance calculation parameter map.
Detailed Description
The tire unbalance amount calibration and measurement method of the present invention will be described in detail with reference to the following embodiments.
The invention relates to a tire unbalance calibration and measurement method, which comprises two parts of tire unbalance calibration and tire unbalance measurement, wherein the two parts are formed by quantity calibration and eccentricity correction, and the method comprises the following steps: what is done in the calibration of the quantities is to calculate a proportionality coefficient between the voltage value measured by the piezoelectric sensor provided on the spindle and the force actually acting on the tire; what is accomplished in the eccentricity correction is that the inherent unbalance of the main shaft and the rim system is measured and calculated by using the proportionality coefficient calculated in the amount calibration; the unbalance of the tire is measured and calculated according to the proportionality coefficient measured in the quantity calibration and the unbalance of the main shaft and the rim system obtained in the eccentricity correction.
Firstly, selecting the specification of a tested tire, conveying the tire to a lubricating station through a conveyor belt, starting a lubricating centering device, stretching a lubricating rod out after the tire is centered, rotating the tire, withdrawing the centering device after the lubrication is finished, starting the conveyor belt, and conveying the tire to a measuring station. After the tire enters a measuring station, a measuring centering device is started, the tire is held in the middle, a lifting platform descends, the tire falls on a lower rim, an upper frame moves in place, a lower chuck is locked, an upper chuck is loosened, after the tire is inflated stably, the tire starts to rotate at the speed of 400rpm along with a main shaft, and a capacity calibration and eccentricity correction program can be operated.
As shown in fig. 1, the quantitative calibration includes: assigning values to the form parameters according to the measurement results; calculating coefficients from the assigned values
Figure A20061012992900081
(ii) a According to the coefficient
Figure A20061012992900082
Calculating coefficients <math><mrow> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mo>=</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>11</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>22</mn> </msub> <mo>-</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>12</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>21</mn> </msub> <mo>;</mo> </mrow></math> According to the coefficient
Figure A20061012992900084
Calculating the proportionality coefficient
Figure A20061012992900085
The measured values of the unbalance of the upper and lower surfaces of the tire are (voltage value, phase) when the tire is not added with the code: (Uu0, Φ u0) and (Ud0, Φ d 0); adding 100 g of weight at the zero-degree position of the upper rim, and measuring the upper unbalance value and the lower unbalance value as (voltage value, phase): (Uu1, Φ u1), (Ud1, Φ d 1); and adding a 100 g weight at the zero-degree position of the lower rim, and measuring the upper unbalance value and the lower unbalance value as (voltage value and phase): (Uu2, Φ u2), (Ud2, Φ d 2).
According to the aboveThe three groups of measurement data are listed as equations, and coefficients are calculated
Let phi 1, phi 2, phi 3, phi 4 beThe corresponding angles are as follows:
<math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>11</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>1</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo></mo> </mrow> <mo>/</mo> <mn>100</mn> <mo>=</mo> <msub> <mrow> <mi>K</mi> <mn>11</mn> </mrow> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>1</mn> </mrow></math>
<math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>12</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>2</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo></mo> </mrow> <mo>/</mo> <mn>100</mn> <mo>=</mo> <msub> <mrow> <mi>K</mi> <mn>12</mn> </mrow> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>2</mn> </mrow></math>
<math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>21</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>1</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo></mo> </mrow> <mo>/</mo> <mn>100</mn> <mo>=</mo> <msub> <mrow> <mi>K</mi> <mn>21</mn> </mrow> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>3</mn> </mrow></math>
<math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>22</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>2</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo></mo> </mrow> <mo>/</mo> <mn>100</mn> <mo>=</mo> <msub> <mrow> <mi>K</mi> <mn>22</mn> </mrow> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>4</mn> </mrow></math>
because: <math><mrow> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mo>=</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>11</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>22</mn> </msub> <mo>-</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>12</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>21</mn> </msub> <mo>=</mo> <msub> <mi>K</mi> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mo>,</mo> </mrow></math> therefore, the method comprises the following steps:
Figure A20061012992900098
Figure A20061012992900099
will be provided with
Figure A200610129929000910
Value decomposition into real and imaginary forms KR、KIThe method comprises the following steps:
Figure A200610129929000912
the mode and angle of the complex number are respectively:
K = K R 2 + K I 2
then it can be obtained:
Figure A200610129929000915
Figure A200610129929000916
Figure A200610129929000917
Figure A200610129929000918
as shown in fig. 2, the eccentricity correction includes: measuring voltage values of 8 groups of upper and lower piezoelectric sensors in the process that the tire rotates for one circle relative to the rim, summing 8 groups of measurement data vectors, and averaging to obtain a voltage value caused by the imbalance of the main shaft and the rim; and converting the voltage value into a corresponding force phase.
Firstly, returning the upper rim and the lower rim to a mechanical zero point; then, with the initial position of the tire as a reference, the relative position of the tire and the upper and lower rims was measured once every 45 degrees, and finally 8 sets of measurement data were obtained (the tire was rotated in a fixed order, and the tire returned to the initial position after 8 measurements). The voltage values of the upper and lower piezoelectric sensors measured each time are respectively set to (U)up)i、(Udp)i(i is 1 to 8), the voltage value caused by the imbalance between the spindle and the rim is:
<math><mrow> <msub> <mover> <mi>U</mi> <mo>&OverBar;</mo> </mover> <mi>uw</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> <munder> <mi>&Sigma;</mi> <mi>i</mi> </munder> <msub> <mrow> <mo>(</mo> <msub> <mover> <mi>U</mi> <mo>&OverBar;</mo> </mover> <mi>up</mi> </msub> <mo>)</mo> </mrow> <mi>i</mi> </msub> </mrow></math>
<math><mrow> <msub> <mover> <mi>U</mi> <mo>&OverBar;</mo> </mover> <mi>dw</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> <munder> <mi>&Sigma;</mi> <mi>i</mi> </munder> <msub> <mrow> <mo>(</mo> <msub> <mover> <mi>U</mi> <mo>&OverBar;</mo> </mover> <mi>dp</mi> </msub> <mo>)</mo> </mrow> <mi>i</mi> </msub> </mrow></math>
the voltage values corresponding to the unbalance amounts of the upper plane and the lower plane of the rim system measured by the piezoelectric sensor are respectively (U)uw,θ1),(Udwθ 2) (unit: volts, degrees).
The unbalance amounts of the upper plane and the lower plane of the main shaft and the rim system at the position of the correction weight are respectively (M)L″,βL″),(MR″,βR"(unit: grams, degrees).
Then there are:
ML″=K1×Uuw+K2×Udw
MR″=K3×Uuw+K4×Udw
wherein:
K1=K1∠α1=K1cosα1+iK1sinα1
K2=K2∠α2=K2cosα2+iK2sinα2
K3=K3∠α3=K3cosα3+iK3sinα3
K4=K4∠α4=K4cosα4+iK4sinα4
therefore, the voltage value (U) corresponding to the unbalance of the upper and lower planesuw,θ1),(Udwθ 2), we can calculate:
<math><mrow> <msub> <mi>M</mi> <msup> <mi>L</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>=</mo> <msqrt> <msup> <mrow> <mo>[</mo> <mi>K</mi> <mn>1</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>uw</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>1</mn> <mo>+</mo> <mi>&theta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>2</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>dw</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>2</mn> <mo>+</mo> <mi>&theta;</mi> <mn>2</mn> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>[</mo> <mi>K</mi> <mn>1</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>uw</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>1</mn> <mo>+</mo> <mi>&theta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>2</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>dw</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>2</mn> <mo>+</mo> <mi>&theta;</mi> <mn>2</mn> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mn>2</mn> </msup> </msqrt> </mrow></math>
<math><mrow> <msub> <mi>&beta;</mi> <msup> <mi>L</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>=</mo> <msup> <mi>tg</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mfrac> <mrow> <mi>K</mi> <mn>1</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>uw</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>1</mn> <mo>+</mo> <mi>&theta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>2</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>dw</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>2</mn> <mo>+</mo> <mi>&theta;</mi> <mn>2</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>k</mi> <mn>1</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>uw</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>1</mn> <mo>+</mo> <mi>&theta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>2</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>dw</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>2</mn> <mo>+</mo> <mi>&theta;</mi> <mn>2</mn> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow></math>
<math><mrow> <msub> <mi>M</mi> <msup> <mi>R</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>=</mo> <msqrt> <msup> <mrow> <mo>[</mo> <mi>K</mi> <mn>3</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>uw</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>3</mn> <mo>+</mo> <mi>&theta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>4</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>dw</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>4</mn> <mo>+</mo> <mi>&theta;</mi> <mn>2</mn> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>[</mo> <mi>K</mi> <mn>3</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>uw</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>3</mn> <mo>+</mo> <mi>&theta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>k</mi> <mn>4</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>dw</mi> </msub> <mi>sin</mi> <mo></mo> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>4</mn> <mo>+</mo> <mi>&theta;</mi> <mn>2</mn> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mn>2</mn> </msup> </msqrt> </mrow></math>
<math><mrow> <msub> <mi>&beta;</mi> <msup> <mi>R</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>=</mo> <msup> <mi>tg</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mfrac> <mrow> <mi>K</mi> <mn>3</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>uw</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>3</mn> <mo>+</mo> <mi>&theta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>4</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>dw</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>4</mn> <mo>+</mo> <mi>&theta;</mi> <mn>2</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>K</mi> <mn>3</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>uw</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>3</mn> <mo>+</mo> <mi>&theta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>4</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>dw</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>4</mn> <mo>+</mo> <mi>&theta;</mi> <mn>2</mn> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow></math>
the unbalance of the rim and spindle system at the weight position is thus obtained as:
ML″=ML″∠βL″,MR″=MR″∠βR″
wherein,is a proportionality coefficient between a voltage value measured by the piezoelectric sensor and a force actually acting on the tire, which is determined in the calibration.
As shown in fig. 3, the measurement of the unbalance amount of the tire includes: converting the voltage value measured by the piezoelectric sensor into corresponding force; calculating the unbalance of the tire at the position of a weight according to the unbalance of the spindle and the rim system obtained by eccentric correction; converting the unbalance of the tire at the position of the weight into the unbalance at the position of the seam allowance, namely a final unbalance value; and calculating the static unbalance and the even unbalance according to the upper unbalance and the lower unbalance.
After the above calibration is completed, the measurement of the amount of unbalance of the tire can be performed, and the measurement is completed in the following stages.
In the first stage, the voltage value measured by the piezoelectric sensor is converted into corresponding force, namely the unbalance of the tire, the rim and the spindle system at the position of the weight.
The voltage values corresponding to the unbalance amounts of the upper plane and the lower plane of the main shaft system of the tire and the rim measured by the piezoelectric sensor are respectively (U)u,β1),(Udβ 2) (unit: volts, degrees).
The unbalance amounts of the upper plane and the lower plane of the tire, the rim and the spindle system at the position of the correction weight are respectively (M)L′,βL′),(MR′,βR') (unit: grams, degrees).
Then there are:
ML′=K1×Uu+K2×Ud
MR′=K3×Uu+K4×Ud
wherein:
K1=K1∠α1=K1cosα1+iK1sinα1
K2=K2∠α2=K2cosα2+iK2sinα2
K3=K3∠α3=K3cosα3+iK3sinα3
K4=K4∠α4=K4cosα4+iK4sinα4
therefore, the voltage value (U) corresponding to the unbalance of the upper and lower planesu,β1),(Udβ 2), one can calculate:
<math><mrow> <msub> <mi>M</mi> <msup> <mi>L</mi> <mo>&prime;</mo> </msup> </msub> <mo>=</mo> <msqrt> <msup> <mrow> <mo>[</mo> <mi>K</mi> <mn>1</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>u</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>1</mn> <mo>+</mo> <mi>&beta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>2</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>d</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>2</mn> <mo>+</mo> <mi>&beta;</mi> <mn>2</mn> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>[</mo> <mi>K</mi> <mn>1</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>u</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>1</mn> <mo>+</mo> <mi>&beta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>2</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>d</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>2</mn> <mo>+</mo> <mi>&beta;</mi> <mn>2</mn> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mn>2</mn> </msup> </msqrt> </mrow></math>
<math><mrow> <msub> <mi>&beta;</mi> <msup> <mi>L</mi> <mo>&prime;</mo> </msup> </msub> <mo>=</mo> <msup> <mi>tg</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mfrac> <mrow> <mi>K</mi> <mn>1</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>u</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>1</mn> <mo>+</mo> <mi>&beta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>2</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>d</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>2</mn> <mo>+</mo> <mi>&beta;</mi> <mn>2</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>K</mi> <mn>1</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>u</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>1</mn> <mo>+</mo> <mi>&beta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>2</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>d</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>2</mn> <mo>+</mo> <mi>&beta;</mi> <mn>2</mn> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow></math>
<math><mrow> <msub> <mi>M</mi> <msup> <mi>R</mi> <mo>&prime;</mo> </msup> </msub> <mo>=</mo> <msqrt> <msup> <mrow> <mo>[</mo> <mi>K</mi> <mn>3</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>u</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>3</mn> <mo>+</mo> <mi>&beta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>4</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>d</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>4</mn> <mo>+</mo> <mi>&beta;</mi> <mn>2</mn> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>[</mo> <mi>K</mi> <mn>3</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>u</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>3</mn> <mo>+</mo> <mi>&beta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>k</mi> <mn>4</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>d</mi> </msub> <mi>sin</mi> <mo></mo> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>4</mn> <mo>+</mo> <mi>&beta;</mi> <mn>2</mn> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mn>2</mn> </msup> </msqrt> </mrow></math>
<math><mrow> <msub> <mi>&beta;</mi> <msup> <mi>R</mi> <mo>&prime;</mo> </msup> </msub> <mo>=</mo> <msup> <mi>tg</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mfrac> <mrow> <mi>K</mi> <mn>3</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>u</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>3</mn> <mo>+</mo> <mi>&beta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>4</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>d</mi> </msub> <mi>sin</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>4</mn> <mo>+</mo> <mi>&beta;</mi> <mn>2</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>K</mi> <mn>3</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>u</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>3</mn> <mo>+</mo> <mi>&beta;</mi> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mn>4</mn> <mo>&times;</mo> <msub> <mi>U</mi> <mi>d</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <mi>&alpha;</mi> <mn>4</mn> <mo>+</mo> <mi>&beta;</mi> <mn>2</mn> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow></math>
the unbalance of the tire, rim and spindle system at the weight position is thus obtained as: mL′=ML′∠βL′,MR′=MR′∠βR′
And in the second stage, calculating the unbalance of the tire at the position of the weight according to the unbalance of the rim and the main shaft system obtained by eccentric correction.
The unbalance amounts of the upper plane and the lower plane of the rim and the main shaft system at the position of the weight are respectively:
(ML″,βL″),(MR″,βR″);
and the unbalance amount of the upper plane and the lower plane of the tire at the position of the weight is set as follows:
(ML,βL),(MR,βR)。
will (M)L′,βL′),(MR′,βR') and (M)L″,βL″),(MR″,βRRespectively subtracting the vectors to obtain the unbalance of the upper surface and the lower surface of the tire.
Unbalance on tire: mL=ML′-ML″
Wherein:
<math><mrow> <msub> <mi>M</mi> <mi>L</mi> </msub> <mo>=</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>M</mi> <msup> <mi>L</mi> <mo>&prime;</mo> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>sin</mi> <mi>&beta;</mi> </mrow> <mi>L</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msub> <mi>M</mi> <msup> <mi>L</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>sin</mi> <mi>&beta;</mi> </mrow> <mi>L</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>M</mi> <msup> <mi>L</mi> <mo>&prime;</mo> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>cos</mi> <mi>&beta;</mi> </mrow> <mi>L</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msub> <mi>M</mi> <msup> <mi>L</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>cos</mi> <mi>&beta;</mi> </mrow> <mi>L</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </msqrt> </mrow></math>
<math><mrow> <msub> <mi>&beta;</mi> <mi>L</mi> </msub> <mo>=</mo> <mi>arctg</mi> <mfrac> <mrow> <mo>(</mo> <msub> <mi>M</mi> <msup> <mi>L</mi> <mo>&prime;</mo> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>sin</mi> <mi>&beta;</mi> </mrow> <mi>L</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msub> <mi>M</mi> <msup> <mi>L</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>sin</mi> <mi>&beta;</mi> </mrow> <mi>L</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>M</mi> <msup> <mi>L</mi> <mo>&prime;</mo> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>cos</mi> <mi>&beta;</mi> </mrow> <mi>L</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msub> <mi>M</mi> <msup> <mi>L</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>cos</mi> <mi>&beta;</mi> </mrow> <mi>L</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>)</mo> </mrow> </mfrac> </mrow></math>
lower tire unbalance amount: mR=MR′-MR″
Wherein:
<math><mrow> <msub> <mi>M</mi> <mi>R</mi> </msub> <mo>=</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>M</mi> <msup> <mi>R</mi> <mo>&prime;</mo> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>sin</mi> <mi>&beta;</mi> </mrow> <mi>R</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msub> <mi>M</mi> <msup> <mi>R</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>sin</mi> <mi>&beta;</mi> </mrow> <mi>R</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>M</mi> <msup> <mi>R</mi> <mo>&prime;</mo> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>cos</mi> <mi>&beta;</mi> </mrow> <mi>R</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msub> <mi>M</mi> <msup> <mi>R</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>cos</mi> <mi>&beta;</mi> </mrow> <mi>R</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </msqrt> </mrow></math>
<math><mrow> <msub> <mi>&beta;</mi> <mi>R</mi> </msub> <mo>=</mo> <mi>arctg</mi> <mfrac> <mrow> <mo>(</mo> <msub> <mi>M</mi> <msup> <mi>R</mi> <mo>&prime;</mo> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>sin</mi> <mi>&beta;</mi> </mrow> <mi>R</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msub> <mi>M</mi> <msup> <mi>R</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>sin</mi> <mi>&beta;</mi> </mrow> <mi>R</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>M</mi> <msup> <mi>R</mi> <mo>&prime;</mo> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>cos</mi> <mi>&beta;</mi> </mrow> <mi>R</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msub> <mi>M</mi> <msup> <mi>R</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </msub> <mo>&times;</mo> <msup> <msub> <mrow> <mi>cos</mi> <mi>&beta;</mi> </mrow> <mi>R</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>)</mo> </mrow> </mfrac> </mrow></math>
and in the third stage, the unbalance amount of the tire in the weight position is converted into the unbalance amount in the seam allowance position, namely the final unbalance amount value.
As shown in fig. 4, the tire dynamic balance calculation parameter map is shown.
Wherein:
r1: the radius of the tire bead plus the radius of the weight; r2: radius at the tire bead;
a: the distance from the upper surface of the tire to the center line of the upper weight; b: upper wheel, 36638and lower wheel, 36638;
c: the distance from the lower surface of the tire to the center line of the lower weight.
Let MLAnd MRThe unbalance amounts of the tire at the weight positions of the upper surface and the lower surface, FLAnd FRIs formed by MLAnd MRResulting unbalanced force, mlAnd mrThe unbalance amounts of the tire at the upper and lower surface seam parts, FlAnd FrIs formed by mlAnd mrThe resulting imbalance forces, the relationship between them is:
FL=MLR1ω2 Fl=MlR2ω2
FR=MRR1ω2 Fr=MrR2ω2
according to the force balance principle of the plane inertia force system, the following equation is established:
FL+FR+Uux+Udx=0
Fl+Fr+Uux+Udx=0
i.e. FL+FR=Fl+Fr
According to the moment balance principle of the plane inertia force system, the following can be obtained:
bFl=(a+b)FL-cFR
bFr=(b+c)FR-aFL
m can be obtainedl、mrAnd ML、MRThe relationship between:
<math><mrow> <msub> <mover> <mi>m</mi> <mo>&OverBar;</mo> </mover> <mi>l</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>a</mi> <mo>+</mo> <mi>b</mi> </mrow> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>-</mo> <mfrac> <mi>c</mi> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>)</mo> </mrow> <mfrac> <msub> <mi>R</mi> <mn>1</mn> </msub> <msub> <mi>R</mi> <mn>2</mn> </msub> </mfrac> </mrow></math>
<math><mrow> <mo>=</mo> <mrow> <mo>(</mo> <mrow> <mo>(</mo> <mi>a</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <mo>&times;</mo> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>-</mo> <mi>c</mi> <mo>&times;</mo> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>)</mo> </mrow> <mfrac> <msub> <mi>R</mi> <mn>1</mn> </msub> <mrow> <mi>b</mi> <mo>.</mo> <msub> <mi>R</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mrow></math>
<math><mrow> <msub> <mover> <mi>m</mi> <mo>&OverBar;</mo> </mover> <mi>r</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>b</mi> <mo>+</mo> <mi>c</mi> </mrow> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>-</mo> <mfrac> <mi>a</mi> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>)</mo> </mrow> <mfrac> <msub> <mi>R</mi> <mn>1</mn> </msub> <msub> <mi>R</mi> <mn>2</mn> </msub> </mfrac> </mrow></math>
<math><mrow> <mo>=</mo> <mrow> <mo>(</mo> <mrow> <mo>(</mo> <mi>b</mi> <mo>+</mo> <mi>c</mi> <mo>)</mo> </mrow> <mo>&times;</mo> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>-</mo> <mi>a</mi> <mo>&times;</mo> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>)</mo> </mrow> <mfrac> <msub> <mi>R</mi> <mn>1</mn> </msub> <mrow> <mi>b</mi> <mo>.</mo> <msub> <mi>R</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mrow></math>
<math><mrow> <mo>=</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>a</mi> <mo>&times;</mo> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mi>b</mi> <mo>+</mo> <mi>c</mi> <mo>)</mo> </mrow> <mo>&times;</mo> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>)</mo> </mrow> <mfrac> <msub> <mi>R</mi> <mn>1</mn> </msub> <mrow> <mi>b</mi> <mo>.</mo> <msub> <mi>R</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mrow></math>
the unbalance amounts of the upper surface and the lower surface of the tire bead position are respectively calculated as (m)l,θl),(mr,θr):
<math><mrow> <msub> <mi>m</mi> <mi>l</mi> </msub> <mo>=</mo> <mfrac> <mrow> <mi>R</mi> <mn>1</mn> </mrow> <mrow> <mi>R</mi> <mn>2</mn> <mo>.</mo> <mi>b</mi> </mrow> </mfrac> <msqrt> <msup> <mrow> <mo>(</mo> <mi>b</mi> <mo>+</mo> <mi>a</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msup> <msub> <mi>M</mi> <mi>L</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>c</mi> <mn>2</mn> </msup> <msup> <msub> <mi>M</mi> <mi>R</mi> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>2</mn> <mi>C</mi> <mrow> <mo>(</mo> <mi>b</mi> <mo>+</mo> <mi>a</mi> <mo>)</mo> </mrow> <msub> <mi>M</mi> <mi>L</mi> </msub> <msub> <mi>M</mi> <mi>R</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <msub> <mi>&phi;</mi> <mi>L</mi> </msub> <mo>-</mo> <msub> <mi>&phi;</mi> <mi>R</mi> </msub> <mo>)</mo> </mrow> </msqrt> </mrow></math>
<math><mrow> <msub> <mi>m</mi> <mi>r</mi> </msub> <mo>=</mo> <mfrac> <mrow> <mi>R</mi> <mn>1</mn> </mrow> <mrow> <mi>R</mi> <mn>2</mn> <mo>.</mo> <mi>b</mi> </mrow> </mfrac> <msqrt> <msup> <mrow> <mo>(</mo> <mi>a</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msup> <msub> <mi>M</mi> <mi>L</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mi>c</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msup> <msub> <mi>M</mi> <mi>R</mi> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>2</mn> <mi>a</mi> <mrow> <mo>(</mo> <mi>b</mi> <mo>+</mo> <mi>c</mi> <mo>)</mo> </mrow> <msub> <mi>M</mi> <mi>L</mi> </msub> <msub> <mi>M</mi> <mi>R</mi> </msub> <mi>cos</mi> <mrow> <mo>(</mo> <msub> <mi>&phi;</mi> <mi>L</mi> </msub> <mo>-</mo> <msub> <mi>&phi;</mi> <mi>R</mi> </msub> <mo>)</mo> </mrow> </msqrt> </mrow></math>
<math><mrow> <msub> <mi>&theta;</mi> <mi>l</mi> </msub> <mo>=</mo> <msup> <mi>tg</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mfrac> <mrow> <mrow> <mo>(</mo> <mi>b</mi> <mo>+</mo> <mi>a</mi> <mo>)</mo> </mrow> <msub> <mi>M</mi> <mi>L</mi> </msub> <mi>sin</mi> <msub> <mi>&phi;</mi> <mi>L</mi> </msub> <mo>-</mo> <msub> <mi>cM</mi> <mi>R</mi> </msub> <msub> <mrow> <mi>sin</mi> <mi>&phi;</mi> </mrow> <mi>R</mi> </msub> </mrow> <mrow> <mrow> <mo>(</mo> <mi>b</mi> <mo>+</mo> <mi>a</mi> <mo>)</mo> </mrow> <msub> <mi>M</mi> <mi>L</mi> </msub> <msub> <mrow> <mi>cos</mi> <mi>&phi;</mi> </mrow> <mi>L</mi> </msub> <mo>-</mo> <msub> <mi>cM</mi> <mi>R</mi> </msub> <msub> <mrow> <mi>cos</mi> <mi>&phi;</mi> </mrow> <mi>R</mi> </msub> </mrow> </mfrac> </mrow></math>
<math><mrow> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> <mo>=</mo> <msup> <mi>tg</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mfrac> <mrow> <mo>-</mo> <mi>a</mi> <msub> <mi>M</mi> <mi>L</mi> </msub> <mi>sin</mi> <msub> <mi>&phi;</mi> <mi>L</mi> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mi>c</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <msub> <mi>M</mi> <mi>R</mi> </msub> <msub> <mrow> <mi>sin</mi> <mi>&phi;</mi> </mrow> <mi>R</mi> </msub> </mrow> <mrow> <mrow> <mo>-</mo> <mi>a</mi> </mrow> <msub> <mi>M</mi> <mi>L</mi> </msub> <msub> <mrow> <mi>cos</mi> <mi>&phi;</mi> </mrow> <mi>L</mi> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mi>c</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <msub> <mi>M</mi> <mi>R</mi> </msub> <msub> <mrow> <mi>cos</mi> <mi>&phi;</mi> </mrow> <mi>R</mi> </msub> </mrow> </mfrac> </mrow></math>
And in the fourth stage, calculating the static unbalance and the even unbalance according to the unbalance of the upper surface and the lower surface.
The static unbalance amount is equal to the product of the vector sum of the upper unbalance amount and the lower unbalance amount and the nominal radius of the wheel rim.
Let ml, mr be: (m)l,θl),(mr,θr) Based on the result, calculate
According to the national standard: wheel \36638.
By using the complex arithmetic theorem, the static unbalance amount can be obtained as follows: <math><mrow> <mover> <mi>S</mi> <mo>&RightArrow;</mo> </mover> <mo>=</mo> <mi>S</mi> <mo>&angle;</mo> <mi>&delta;</mi> <mo>.</mo> </mrow></math>
wherein, <math><mrow> <mover> <mi>S</mi> <mo>&RightArrow;</mo> </mover> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mover> <mi>m</mi> <mo>&RightArrow;</mo> </mover> <mi>l</mi> </msub> <mo>+</mo> <msub> <mover> <mi>m</mi> <mo>&RightArrow;</mo> </mover> <mi>r</mi> </msub> <mo>)</mo> </mrow> <mo>&times;</mo> <mi>R</mi> <mo>.</mo> </mrow></math>
real part: ss is ml×cosθl+mr×cosθr
Imaginary part: sx ═ ml×sinθl+mr×sinθr
S12=Ss×Ss+Sx×Sx
<math><mrow> <mi>S</mi> <mo>=</mo> <msqrt> <mi>S</mi> <mn>1</mn> </msqrt> <mo>&times;</mo> <mi>R</mi> </mrow></math>
δ=arctg(Sx/Ss)
The even unbalance amount is equal to the product of the vector difference of the upper unbalance amount and the lower unbalance amount and the nominal rim radius and the rim width.
Let ml, mr be: (m)l,θl),(mr,θr) (ii) a Wheel \36638, nominal radius R ═ (D/2) × 2.54+1.27, where D is the rim diameter in inches; the rim Width is Width in centimeters. The even imbalance is calculated as follows: <math><mrow> <mover> <mi>C</mi> <mo>&RightArrow;</mo> </mover> <mo>=</mo> <mi>C</mi> <mo>&angle;</mo> <mi>&sigma;</mi> <mo>.</mo> </mrow></math>
wherein, <math><mrow> <mover> <mi>C</mi> <mo>&RightArrow;</mo> </mover> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mover> <mi>m</mi> <mo>&RightArrow;</mo> </mover> <mi>l</mi> </msub> <mo>-</mo> <msub> <mover> <mi>m</mi> <mo>&RightArrow;</mo> </mover> <mi>r</mi> </msub> <mo>)</mo> </mrow> <mo>&times;</mo> <mi>R</mi> <mo>&times;</mo> <mi>Width</mi> <mo>.</mo> </mrow></math>
real part: cs ═ ml×cosθl-mr×cosθr
Imaginary part: cx ═ ml×sinθl-mr×sinθr
C12=Cs×Cs+Cx×Cx
<math><mrow> <mi>C</mi> <mo>=</mo> <msqrt> <mi>C</mi> <mn>1</mn> </msqrt> <mo>&times;</mo> <mi>R</mi> <mo>&times;</mo> <mi>Width</mi> </mrow></math>
σ=arctg(Cx/Cs)
After the stages are completed, the main shaft rotates to a designated position according to the calculated marking angle and stops rotating, the tire is deflated, the upper chuck is locked, the lower chuck is loosened, the upper frame returns to the original point, the lifting platform is lifted, and the conveyor belt is started to convey the tire to the marking station. The marking station pin stretches out, the marking motor is started to drive the marking mechanism to move in place, the marking head corresponding to the tire grade stretches out, after marking is completed, the marking mechanism returns to the original position, the pin is retracted, and the conveyor belt is started to convey tires to the output station. The output cylinder of the output station is retracted, and after the tire falls down, the output cylinder extends out, and the action is finished.

Claims (11)

1. A tire unbalance calibration and measurement method is characterized by comprising two parts of tire unbalance calibration and tire unbalance measurement, wherein the two parts are formed by quantity calibration and eccentricity correction, and the two parts are as follows: what is done in the calibration of the quantities is to calculate a proportionality coefficient between the voltage value measured by the piezoelectric sensor provided on the spindle and the force actually acting on the tire; what is accomplished in the eccentricity correction is that the inherent unbalance of the main shaft and the rim system is measured and calculated by using the proportionality coefficient calculated in the amount calibration; the unbalance of the tire is measured and calculated according to the proportionality coefficient measured in the quantity calibration and the unbalance of the main shaft and the rim system obtained in the eccentricity correction.
2. The method for calibrating and measuring the amount of unbalance of a tire as claimed in claim 1, wherein said amount calibration comprises: assigning values to the form parameters according to the measurement results; calculating coefficients from the assigned values
Figure A2006101299290002C1
According to the coefficientCalculating coefficients <math><mrow> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mo>=</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>11</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>22</mn> </msub> <mo>-</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>12</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>21</mn> </msub> <mo>;</mo> </mrow></math> According to the coefficient
Figure A2006101299290002C4
Calculating the proportionality coefficient
Figure A2006101299290002C5
Figure A2006101299290002C6
Figure A2006101299290002C7
Figure A2006101299290002C8
3. The method for calibrating and measuring the amount of unbalance of a tire according to claim 2, wherein the step of assigning values to form parameters according to the measurement results comprises the steps of, when no weight is added to the tire, measuring the following voltage values on the tire: uu0, Φ u0, Ud0, Φ d 0; when weights are added on the upper surface of the tire, the voltage values of the upper surface and the lower surface of the tire are respectively measured as follows: uu1, Φ u1, Ud1, Φ d 1; when weights are added below the tire, the voltage values above and below the tire are respectively measured: uu2, Φ u2 and Ud2, Φ d 2.
4. The method for calibrating and measuring the amount of unbalance of a tire as claimed in claim 2, wherein the coefficient is calculated based on the assigned valueComprises the following steps: calculate out <math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>11</mn> </msub> <mo>=</mo> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>1</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo>/</mo> <mn>100</mn> <mo>=</mo> <mi>K</mi> <mn>1</mn> <msub> <mn>1</mn> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>1</mn> </mrow></math> A stage (2); calculate out <math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>12</mn> </msub> <mo>=</mo> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>2</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>u</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;u</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo>/</mo> <mn>100</mn> <mo>=</mo> <mi>K</mi> <mn>1</mn> <msub> <mn>2</mn> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>2</mn> </mrow></math> A stage (2); calculate out <math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>21</mn> </msub> <mo>=</mo> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>1</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo>/</mo> <mn>100</mn> <mo>=</mo> <mi>K</mi> <mn>2</mn> <msub> <mn>1</mn> <mrow> <mo>&angle;</mo> <mi>&phi;</mi> </mrow> </msub> <mn>3</mn> </mrow></math> A stage (2); calculate out <math><mrow> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>22</mn> </msub> <mo>=</mo> <mo>(</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>2</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>U</mi> <mrow> <mi>d</mi> <mn>0</mn> <mo>&angle;</mo> <mi>&Phi;d</mi> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo>/</mo> <mn>100</mn> <mo>=</mo> <mi>K</mi> <mn>2</mn> <msub> <mn>2</mn> <mrow> <mo>&angle;</mo> <mi>&Phi;</mi> </mrow> </msub> <mn>4</mn> </mrow></math> And (3) a stage of (a).
5. The method for calibrating and measuring the amount of unbalance of a tire as claimed in claim 2, wherein the coefficient of dependence is
Figure A2006101299290002C16
Calculating coefficients
Figure A2006101299290002C17
The method comprises the following steps: <math><mrow> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mo>=</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>11</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>22</mn> </msub> <mo>-</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>12</mn> </msub> <mo>&times;</mo> <msub> <mover> <mi>K</mi> <mo>&RightArrow;</mo> </mover> <mn>21</mn> </msub> <mo>.</mo> </mrow></math>
6. the method for calibrating and measuring the amount of unbalance of a tire as claimed in claim 2, wherein the coefficient of dependence isCalculating the proportionality coefficient
Figure A2006101299290003C2
Figure A2006101299290003C3
Figure A2006101299290003C4
Figure A2006101299290003C5
The method comprises the following steps:
Figure A2006101299290003C6
Figure A2006101299290003C7
Figure A2006101299290003C8
Figure A2006101299290003C9
7. the method for calibrating and measuring the amount of tire unbalance as claimed in claim 1, wherein said eccentricity correction comprises: measuring voltage values of 8 groups of upper and lower piezoelectric sensors in the process that the tire rotates for one circle relative to the rim, summing 8 groups of measurement data vectors, and averaging to obtain a voltage value caused by the imbalance of the main shaft and the rim; and converting the voltage value into a corresponding force phase.
8. The method for calibrating and measuring the amount of unbalance of a tire as claimed in claim 7, wherein the step of converting the voltage value into the corresponding force is to set the voltage value corresponding to the amount of unbalance between the upper and lower planes of the rim system and the spindle measured by the piezoelectric sensor; setting the unbalance amount of the upper plane and the lower plane of the main shaft and the rim system at the position of the correction weight; the corresponding forces are thus calculated as:
<math><mrow> <msup> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>=</mo> <mover> <mi>K</mi> <mo>&OverBar;</mo> </mover> <mn>1</mn> <mo>&times;</mo> <msub> <mover> <mi>U</mi> <mo>&OverBar;</mo> </mover> <mi>uw</mi> </msub> <mo>+</mo> <mover> <mi>K</mi> <mo>&OverBar;</mo> </mover> <mn>2</mn> <mo>&times;</mo> <msub> <mover> <mi>U</mi> <mo>&OverBar;</mo> </mover> <mi>dw</mi> </msub> </mrow></math>
<math><mrow> <msup> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>=</mo> <mover> <mi>K</mi> <mo>&OverBar;</mo> </mover> <mn>3</mn> <mo>&times;</mo> <msub> <mover> <mi>U</mi> <mo>&OverBar;</mo> </mover> <mi>uw</mi> </msub> <mo>+</mo> <mover> <mi>K</mi> <mo>&OverBar;</mo> </mover> <mn>4</mn> <mo>&times;</mo> <msub> <mover> <mi>U</mi> <mo>&OverBar;</mo> </mover> <mi>dw</mi> </msub> <mo>.</mo> </mrow></math>
9. the method for calibrating and measuring the amount of tire unbalance as claimed in claim 1, wherein said measuring the amount of tire unbalance comprises: converting the voltage value measured by the piezoelectric sensor into corresponding force; calculating the unbalance of the tire at the position of a weight according to the unbalance of the spindle and the rim system obtained by eccentric correction; converting the unbalance of the tire at the position of the weight into the unbalance at the position of the seam allowance, namely a final unbalance value; and calculating the static unbalance and the even unbalance according to the upper unbalance and the lower unbalance.
10. The method for calibrating and measuring the unbalance amount of the tire as claimed in claim 9, wherein the step of calculating the unbalance amount of the tire at the weight position according to the unbalance amount of the spindle and rim system obtained by the eccentricity correction comprises the steps of firstly setting the unbalance amounts of the upper and lower planes of the spindle and rim system at the weight position as follows: mL″,βL"and MR″,βR"; and then the measured unbalance amounts of the upper plane and the lower plane of the tire at the position of the weight are respectively set as: mL,βLAnd MR,βR(ii) a Will ML′,βL' and MR′,βR' and ML″,βL"and MR″,βR"the respective vectors are subtracted to obtain the upper and lower unbalance of the tire itself:
unbalance on tire: <math><mrow> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>=</mo> <msup> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msup> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>,</mo> </mrow></math>
lower tire unbalance amount: <math><mrow> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>=</mo> <msup> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msup> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mo>.</mo> </mrow></math>
11. the method as claimed in claim 9, wherein the step of converting the unbalance of the tire at the weight position into the unbalance at the bead position, i.e. the final unbalance value, comprises calculating the unbalance of the upper and lower surfaces of the bead position of the tire as m respectively according to the measurementl,θlAnd mr,θrWherein:
<math><mrow> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>l</mi> </msub> <mo>=</mo> <mo>(</mo> <mfrac> <mrow> <mi>a</mi> <mo>+</mo> <mi>b</mi> </mrow> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>-</mo> <mfrac> <mi>c</mi> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>)</mo> <mfrac> <msub> <mi>R</mi> <mn>1</mn> </msub> <msub> <mi>R</mi> <mn>2</mn> </msub> </mfrac> <mo>,</mo> </mrow></math> <math><mrow> <msub> <mover> <mi>m</mi> <mo>&OverBar;</mo> </mover> <mi>r</mi> </msub> <mo>=</mo> <mo>(</mo> <mfrac> <mrow> <mi>b</mi> <mo>+</mo> <mi>c</mi> </mrow> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>R</mi> </msub> <mo>-</mo> <mfrac> <mi>a</mi> <mi>b</mi> </mfrac> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>L</mi> </msub> <mo>)</mo> <mfrac> <msub> <mi>R</mi> <mn>1</mn> </msub> <msub> <mi>R</mi> <mn>2</mn> </msub> </mfrac> <mo>.</mo> </mrow></math>
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