CN107882950A - A kind of involute profile correction method of harmonic drive - Google Patents

A kind of involute profile correction method of harmonic drive Download PDF

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CN107882950A
CN107882950A CN201711032949.XA CN201711032949A CN107882950A CN 107882950 A CN107882950 A CN 107882950A CN 201711032949 A CN201711032949 A CN 201711032949A CN 107882950 A CN107882950 A CN 107882950A
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CN107882950B (en
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陈盛花
张杨
郑建林
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Guomao precision transmission (Changzhou) Co.,Ltd.
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Guangzhou Start To Sail Industrial Robot Co
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    • F—MECHANICAL ENGINEERING; LIGHTING; HEATING; WEAPONS; BLASTING
    • F16—ENGINEERING ELEMENTS AND UNITS; GENERAL MEASURES FOR PRODUCING AND MAINTAINING EFFECTIVE FUNCTIONING OF MACHINES OR INSTALLATIONS; THERMAL INSULATION IN GENERAL
    • F16H—GEARING
    • F16H55/00—Elements with teeth or friction surfaces for conveying motion; Worms, pulleys or sheaves for gearing mechanisms
    • F16H55/02—Toothed members; Worms
    • F16H55/08—Profiling
    • F16H55/0806—Involute profile
    • F—MECHANICAL ENGINEERING; LIGHTING; HEATING; WEAPONS; BLASTING
    • F16—ENGINEERING ELEMENTS AND UNITS; GENERAL MEASURES FOR PRODUCING AND MAINTAINING EFFECTIVE FUNCTIONING OF MACHINES OR INSTALLATIONS; THERMAL INSULATION IN GENERAL
    • F16H—GEARING
    • F16H49/00—Other gearings
    • F16H49/001—Wave gearings, e.g. harmonic drive transmissions
    • F—MECHANICAL ENGINEERING; LIGHTING; HEATING; WEAPONS; BLASTING
    • F16—ENGINEERING ELEMENTS AND UNITS; GENERAL MEASURES FOR PRODUCING AND MAINTAINING EFFECTIVE FUNCTIONING OF MACHINES OR INSTALLATIONS; THERMAL INSULATION IN GENERAL
    • F16H—GEARING
    • F16H55/00—Elements with teeth or friction surfaces for conveying motion; Worms, pulleys or sheaves for gearing mechanisms
    • F16H55/02—Toothed members; Worms
    • F16H55/08—Profiling
    • F16H55/088—Profiling with corrections on tip or foot of the teeth, e.g. addendum relief for better approach contact

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  • Engineering & Computer Science (AREA)
  • General Engineering & Computer Science (AREA)
  • Mechanical Engineering (AREA)
  • Other Investigation Or Analysis Of Materials By Electrical Means (AREA)
  • Pharmaceuticals Containing Other Organic And Inorganic Compounds (AREA)
  • Gears, Cams (AREA)

Abstract

The invention discloses a kind of involute profile correction method of harmonic drive, point B is the point of contact of modification curve and involute flank, point E is the intersection point of modification curve and involute teeth tip circle, establishes coordinate system XOY by axis of abscissas of ray OA, coordinate system X is established by axis of ordinates of ray OC1OY1.By determining point B and point E in coordinate system X1OY1In tangent slope at point B of coordinate and involute, determine flexbile gear tooth central point approximate motion trajectory curve equation, determine modification curve in coordinate system X1OY1In the step such as parametric equation after, Simultaneous Equations solve each parameter, and are converted to parametric equation of the modification curve under coordinate system XOY by coordinate system.This method is to carry out correction of the flank shape to harmonic drive involute profile based on harmonic drive kinematical theory, it is set both to can guarantee that processing technology and economy, and can improves meshing performance, the advantages of combining various harmonic wave tooth forms, can be widely applied to harmonic gear technical field.

Description

A kind of involute profile correction method of harmonic drive
Technical field
The present invention relates to harmonic gear technical field, the involute profile correction method of more particularly to a kind of harmonic drive.
Background technology
Involute profile has the advantages that good transfer performance, technique sexal maturity, processing and manufacturing cutter are simple, extensively should For in gear drive.And steel wheel is nibbled out because in Harmonic Gears, flexbile gear is frequently engaging-in, path of action is similar to put Line, cause common involute gear meshing performance in harmonic drive poor, and cusp contact phenomena in engagement process be present, no Beneficial to the formation of oil film.Traditional involute profile repairing type method is established on the basis of willis gears engage theorem substantially, is fitted For dead axle transmission, Harmonic Gears are not suitable for.
The content of the invention
The technical problems to be solved by the invention are to provide a kind of involute profile correction method of harmonic drive, using this Method carries out profile modification and axial modification to involute profile, and the flexbile gear gear teeth can overcome the disadvantages that involute gear passes for harmonic gear Dynamic deficiency, it can increase while the number of teeth of engagement, can make that meshing backlass is smaller, can realize continuous transmission, can guarantee that processing work Skill and economy, and can improve meshing performance, the advantages of combining various harmonic wave tooth forms, so as to improve the steady of transmission Property.
Technical scheme used by solve above-mentioned technical problem：
A kind of involute profile correction method of harmonic drive, correction of the flank shape is carried out to involute profile along modification curve, point B is The point of contact of modification curve and involute flank, point E are the intersection point of modification curve and involute teeth tip circle, and point O is the basic circle center of circle, with Ray OA is that abscissa establishes coordinate system XOY, and coordinate system X is established by axis of ordinates of ray OC1OY1, point A is the flank of tooth and basic circle Intersection point, point C is the intersection point of gear teeth center line and outside circle；The step of solving modification curve is as follows：
1-1, in coordinate system X1OY1In, it is determined that point B and point E coordinate is respectively (x1B,y1B)、(x1E,y1E), and gradually open Tangent slope k of the line at point B1B；
1-2, in coordinate system X1OY1In, flexbile gear tooth central point approximate motion geometric locus is determined according to rack approximation method principle Equation is
Wherein, m be flexbile gear modulus, z2For the number of teeth of flexbile gear, k is the slope of any point on the curve, and the ψ of η=2, ψ is soft Angle between the radius vector of the intersection point of gear teeth center line and neutral line curve and the major axis of harmonic oscillator；
1-3, flexbile gear pond central point approximate motion trajectory curve equation is translated, rotated and scaled acquisition while is passed through Point B and point E modification curve, and determine modification curve in coordinate system X1OY1In parametric equation be
Wherein, λ, χ, τ be modification curve parameter value, η ∈ [0, π]；
1-4, Simultaneous Equations can obtain the parameter values such as λ in modification curve, χ, τ, and modification curve is corresponding at the point B ηBWith at point E corresponding to ηE,
1-5, coordinate system conversion is carried out, obtaining parametric equation of the modification curve under coordinate system XOY is
Wherein, η ∈ [ηB,ηE]。
Further, coordinate system XOY and X1OY1Between transition matrix be
Wherein,θ0=tan α0-α0, α0For pressure angle of graduated circle.
Further, determine point B in coordinate system X in the step 1-11OY1In coordinate step it is as follows,
The parametric equation of 1-1-1, involute in coordinate system XOY is
Equation of the outside circle in coordinate system XOY be
By above-mentioned two equations simultaneousness can invocation point K involute roll angle φKFor
1-1-2, the coordinate in coordinate system XOY for determining point B, arc KA length are
Similarly, arc BA and arc QA length are respectively
Therefore arc BK length is
I.e.
So roll angle φs of the point B in involuteBFor
Coordinates of the point B in coordinate system XOY, which can be obtained, is
1-1-3, determine point B in coordinate system X1OY1In coordinate be
Further, determine point E in coordinate system X in the step 1-11OY1In coordinate method it is as follows, according to point K rolling Dynamic angle must can exhibit angle θKFor
θK=φK-arctanφk
Therefore
βK=β0+θ0-θK
So point E is in coordinate system X1OY1In coordinate be
Further, according to the step 1-1-3 midpoints B in coordinate system X1OY1In coordinate can draw involute at point B Slope be
Further, the position of the E points is determined by maximum profiling quantity at tooth top, the maximum profiling quantity and flexbile gear wheel The modulus m of tooth is relevant, and span is (0.05m, 0.4m).
Beneficial effect：This method is to carry out correction of the flank shape to harmonic drive involute profile based on harmonic drive kinematical theory, The path curves of flexbile gear tooth central point are obtained by Harmonic Gears kinematic geometry theory, the movement locus is carried out Modification curve after scaling and rotation as harmonic wave involute profile, the modification curve intersect with the end points of flexbile gear tooth top, simultaneously It is again tangent with the flexbile gear flank of tooth, so that the flank of tooth of flexbile gear smoothly transits.Correction of the flank shape, flexbile gear wheel are carried out to flexbile gear flank profil using this method Tooth can overcome the disadvantages that involute gear be used for Harmonic Gears deficiency, can increase while engage the number of teeth, meshing backlass can be made more It is small, continuous transmission can be realized, both can guarantee that processing technology and economy, and can improves meshing performance, can be widely applied to Harmonic gear technical field.
Brief description of the drawings
Fig. 1 is tooth top tip relief schematic diagram；
Fig. 2 is correction of the flank shape part and modification curve schematic diagram on the right side of flexbile gear gear.
Embodiment
Correction of the flank shape is carried out to involute profile along modification curve in this method, it is bent to correction of the flank shape of the present invention with reference to Fig. 1 and Fig. 2 The method for solving of line is described further.
Such as Fig. 1, dotted portion is exactly modification curve.Such as Fig. 2, to carry out correction of the flank shape to the involute profile on the right side of the flexbile gear gear teeth Exemplified by illustrate, point B is the intersection point of modification curve and involute flank, and point E is the intersection point of modification curve and outside circle, and Modification curve and involute flank are tangent at B points.The position of the E points passes through maximum profiling quantity Δ at tooth topmaxIt is determined that The maximum profiling quantity ΔmaxIt is relevant with the modulus m of the flexbile gear gear teeth, maximum profiling quantity ΔmaxSpan for (0.05m, 0.4m)。
Point K is the intersection point of outside circle and the flank of tooth before correction of the flank shape, and point Q is the intersection point of reference circle and the flank of tooth, and point A is the flank of tooth and basic circle Intersection point, point O is the basic circle center of circle, and point C is the intersection point of gear teeth center line and outside circle.The symbol description being hereinafter related to such as table 1 It is shown.
Using point O as the origin of coordinates, coordinate system XOY is established by axis of abscissas of ray OA, ray OA is X in coordinate system XOY The positive coordinate axle of axle, coordinate system X is established by the longitudinal axis of ray OC1OY1, ray OC is coordinate system X1OY1Middle Y1The positive coordinate axle of axle. Coordinate system XOY and X1OY1Between transition matrix be
Wherein,θ0=tan α0-α0。
The step of solving modification curve is as follows：
1-1, determine point B and point E in coordinate system X1OY1In coordinate be respectively (x1B,y1B)、(x1E,y1E)。
If involute roll angle is φ, therefore parametric equation of the involute in coordinate system XOY is
Wherein, base radius
Equation of the outside circle in coordinate system XOY be
Wherein, radius of addendum
Simultaneous involute equation and tooth top equation of a circle can obtain
So as to point K roll angle φKFor
So arc KA length is
Similarly, arc BA and arc QA length are respectively
Therefore, the length that can obtain arc BK and arc QK is respectively
Can obtain arc BK length by Fig. 2 is
So
Therefore
Coordinate in point B coordinate systems XOY is：
Therefore, point B is in coordinate system X1OY1In coordinate be
Meanwhile the coordinate derivation to point B can obtain tangent slope of the involute at point B and be
In addition, the roll angle φ according to point KkPoint K exhibition angle θ can be drawnKFor
θK=φK-arctanφk
Therefore can obtain
βK=β0+θ0-θK
So point E is in coordinate system X1OY1In coordinate be
In X1OY1In coordinate system, it can be obtained according to rack approximation method principle, flexbile gear tooth central point approximate motion geometric locus side Cheng Wei
Wherein, the ψ of η=2, ψ are the radius vector of intersection point and the major axis of harmonic oscillator of flexbile gear tooth center line and neutral line curve Between angle.
Above-mentioned approximate motion trajectory curve equation is translated, rotated and scaled, is obtained simultaneously by point B and point E Modification curve, it is in coordinate system X1OY1In parametric equation be
Wherein, η ∈ [0, π].
Therefore, following five yuan of Nonlinear System of Equations be can obtain
Parameter lambda in modification curve, χ, τ value are can be obtained by by solving above-mentioned equation group, and correction of the flank shape can be obtained η on curve corresponding to point BBWith the η corresponding to point EE.By the conversion of coordinate system, modification curve can be obtained in coordinate system Parametric equation under XOY is：
Wherein, η ∈ [ηB,ηE]。
Embodiment one：
Specifically calculated with reference to the modification curve of a specific flexbile gear gear.Major parameter is chosen as follows： Flexbile gear number of teeth z=60, modulus m=1.3, pressure angle α0=20 °, addendum coefficientThen β0=pi/2 z=0.0262, θ0 =tan α0-α0=0.0149.If the maximum profiling quantity and correction of the flank shape length of harmonic wave involute profile correction of the flank shape are respectively Δmax=0.2m =0.26,Parametric equation of the involute in coordinate system XOY be
Wherein, base radius
Equation in outside circle coordinate system XOY is
Wherein, radius of addendum
Simultaneous involute and tooth top equation of a circle can obtain
So as to invocation point K roll angle φKFor
Therefore, arc KA length is
The length that arc BA and arc QA can similarly be obtained is respectively
Therefore, the length that can obtain arc BK and arc QK is respectively
Can obtain arc BK length by Fig. 2 is
So
Therefore
So in coordinate system X1OY1Midpoint B coordinate is
And in coordinate system X1OY1In, tangent slope of the involute at point B is
In addition, the roll angle φ according to point KkPoint K exhibition angle θ can be drawnKFor
θK=φK-arctanφk=0.0255
Therefore can obtain
βK=β0+θ0-θK=0.0156
So point E is in coordinate system X1OY1Coordinate be
By x1B,y1B,k1B,x1E,y1E, m value five yuan of Nonlinear System of Equations of substitution, obtain nonlinear equation
Therefore can obtain
λ=0.2352, χ=1.6279, τ=37.7601, ηB=1.1795, ηE=2.3136 thus obtain modification curve and sitting Mark system X1OY1In parametric equation be
By arranging：
Wherein, η ∈ [1.1795,2.3136].
Table 1
Embodiments of the present invention are explained in detail above in association with accompanying drawing, but the invention is not restricted to above-mentioned embodiment party Formula, can also be before present inventive concept not be departed from the technical field those of ordinary skill possessed knowledge Put that various changes can be made.

Claims (6)

1. a kind of involute profile correction method of harmonic drive, correction of the flank shape, its feature are carried out to involute profile along modification curve It is：
Point B is the point of contact of modification curve and involute flank, and point E is the intersection point of modification curve and involute teeth tip circle, and point O is base The round heart, point A are the intersection point of the flank of tooth and basic circle, and point C is the intersection point of gear teeth center line and outside circle, using ray OA as axis of abscissas Coordinate system XOY is established, coordinate system X is established by axis of ordinates of ray OC1OY1；
The step of solving modification curve is as follows：
1-1, in coordinate system X1OY1In, it is determined that point B and point E coordinate is respectively (x1B,y1B)、(x1E,y1E), and involute exists Tangent slope k at point B1B；
1-2, in coordinate system X1OY1In, flexbile gear tooth central point approximate motion trajectory curve equation is determined according to rack approximation method principle For
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mi>x</mi> <mo>=</mo> <mn>0.5</mn> <mi>m</mi> <mrow> <mo>(</mo> <mi>&amp;eta;</mi> <mo>-</mo> <mi>k</mi> <mi> </mi> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>&amp;eta;</mi> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mi>y</mi> <mo>=</mo> <mi>k</mi> <mi>m</mi> <mi> </mi> <mi>cos</mi> <mi>&amp;eta;</mi> <mo>+</mo> <mn>0.5</mn> <msub> <mi>mz</mi> <mn>2</mn> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced>
Wherein, m be flexbile gear modulus, z2For the number of teeth of flexbile gear, k is the slope of any point on the curve, and the ψ of η=2, ψ is flexbile gear tooth Angle between the radius vector of the intersection point of center line and neutral line curve and the major axis of harmonic oscillator；
1-3, by flexbile gear pond central point approximate motion trajectory curve equation translated, rotated and scaled acquisition simultaneously by point B With point E modification curve, and determine modification curve in coordinate system X1OY1In parametric equation be
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>x</mi> <mn>1</mn> </msub> <mo>=</mo> <mo>-</mo> <mn>0.5</mn> <mi>&amp;pi;</mi> <mi>m</mi> <mi>&amp;lambda;</mi> <mo>-</mo> <mn>0.5</mn> <mi>m</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>&amp;lambda;</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>&amp;eta;</mi> <mo>-</mo> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>&amp;eta;</mi> <mo>)</mo> </mrow> <mo>+</mo> <mi>&amp;chi;</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>y</mi> <mn>1</mn> </msub> <mo>=</mo> <mn>2</mn> <mi>m</mi> <mi>&amp;lambda;</mi> <mo>-</mo> <mi>m</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>&amp;lambda;</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>+</mo> <mi>cos</mi> <mi>&amp;eta;</mi> <mo>)</mo> </mrow> <mo>+</mo> <mi>&amp;tau;</mi> </mrow> </mtd> </mtr> </mtable> </mfenced>
Wherein, λ, χ, τ be modification curve parameter value, η ∈ [0, π]；
1-4, Simultaneous Equations can obtain the parameter values such as λ in modification curve, χ, τ, and modification curve η corresponding at the point BBWith η corresponding at the point EE,
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mn>2</mn> <msub> <mi>sin&amp;eta;</mi> <mi>B</mi> </msub> </mrow> <mrow> <mo>-</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>cos&amp;eta;</mi> <mi>B</mi> </msub> </mrow> </mfrac> <mo>=</mo> <msub> <mi>k</mi> <mrow> <mn>1</mn> <mi>B</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>-</mo> <mn>0.5</mn> <mi>&amp;pi;</mi> <mi>m</mi> <mi>&amp;lambda;</mi> <mo>-</mo> <mn>0.5</mn> <mi>m</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>&amp;lambda;</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>&amp;eta;</mi> <mi>B</mi> </msub> <mo>-</mo> <msub> <mi>sin&amp;eta;</mi> <mi>B</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>&amp;chi;</mi> <mo>=</mo> <msub> <mi>x</mi> <mrow> <mn>1</mn> <mi>B</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>2</mn> <mi>m</mi> <mi>&amp;lambda;</mi> <mo>-</mo> <mi>m</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>&amp;lambda;</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>cos&amp;eta;</mi> <mi>B</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>&amp;tau;</mi> <mo>=</mo> <msub> <mi>y</mi> <mrow> <mn>1</mn> <mi>B</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>-</mo> <mn>0.5</mn> <mi>&amp;pi;</mi> <mi>m</mi> <mi>&amp;lambda;</mi> <mo>-</mo> <mn>0.5</mn> <mi>m</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>&amp;lambda;</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>&amp;eta;</mi> <mi>E</mi> </msub> <mo>-</mo> <msub> <mi>sin&amp;eta;</mi> <mi>E</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>&amp;chi;</mi> <mo>=</mo> <msub> <mi>x</mi> <mrow> <mn>1</mn> <mi>E</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>2</mn> <mi>m</mi> <mi>&amp;lambda;</mi> <mo>-</mo> <mi>m</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>&amp;lambda;</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>cos&amp;eta;</mi> <mi>E</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>&amp;tau;</mi> <mo>=</mo> <msub> <mi>y</mi> <mrow> <mn>1</mn> <mi>E</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced>
1-5, coordinate system conversion is carried out, obtaining parametric equation of the modification curve under coordinate system XOY is
<mrow> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>x</mi> </mtd> </mtr> <mtr> <mtd> <mi>y</mi> </mtd> </mtr> </mtable> </mfenced> <mo>=</mo> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> <mtd> <mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>-</mo> <mi>cos</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> <mtd> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mo>-</mo> <mn>0.5</mn> <mi>&amp;pi;</mi> <mi>m</mi> <mi>&amp;lambda;</mi> <mo>-</mo> <mn>0.5</mn> <mi>m</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>&amp;lambda;</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>&amp;eta;</mi> <mo>-</mo> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>&amp;eta;</mi> <mo>)</mo> </mrow> <mo>+</mo> <mi>&amp;chi;</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>2</mn> <mi>m</mi> <mi>&amp;lambda;</mi> <mo>-</mo> <mi>m</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>&amp;lambda;</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>+</mo> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>&amp;eta;</mi> <mo>)</mo> </mrow> <mo>+</mo> <mi>&amp;tau;</mi> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
Wherein, η ∈ [ηB,ηE]。
A kind of 2. involute profile correction method of harmonic drive according to claim 1, it is characterised in that：Coordinate system XOY and X1OY1Between transition matrix be
<mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> <mtd> <mrow> <mo>-</mo> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> <mtd> <mrow> <mi>sin</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced>
Wherein,θ0=tan α0-α0, α0For pressure angle of graduated circle.
A kind of 3. involute profile correction method of harmonic drive according to claim 2, it is characterised in that：The step Determine point B in coordinate system X in 1-11OY1In coordinate step it is as follows,
The parametric equation of 1-1-1, involute in coordinate system XOY is
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mi>x</mi> <mo>=</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>&amp;phi;</mi> <mo>+</mo> <mi>&amp;phi;</mi> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>&amp;phi;</mi> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mi>y</mi> <mo>=</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>&amp;phi;</mi> <mo>-</mo> <mi>&amp;phi;</mi> <mi>cos</mi> <mi>&amp;phi;</mi> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced>
Equation of the outside circle in coordinate system XOY be
<mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>y</mi> <mn>2</mn> </msup> <mo>=</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> </mrow>
By above-mentioned two equations simultaneousness can invocation point K involute roll angle φKFor
<mrow> <msub> <mi>&amp;phi;</mi> <mi>K</mi> </msub> <mo>=</mo> <mfrac> <msqrt> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>-</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> </msqrt> <msub> <mi>r</mi> <mi>b</mi> </msub> </mfrac> </mrow>
1-1-2, the coordinate in coordinate system XOY for determining point B, arc KA length are
<mrow> <msub> <mi>L</mi> <mrow> <mi>K</mi> <mi>A</mi> </mrow> </msub> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <msub> <mi>&amp;phi;</mi> <mi>K</mi> </msub> </msubsup> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>d</mi> <mi>x</mi> </mrow> <mrow> <mi>d</mi> <mi>&amp;phi;</mi> </mrow> </mfrac> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>d</mi> <mi>y</mi> </mrow> <mrow> <mi>d</mi> <mi>&amp;phi;</mi> </mrow> </mfrac> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </msqrt> <mi>d</mi> <mi>&amp;phi;</mi> <mo>=</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <msub> <mi>&amp;phi;</mi> <mi>K</mi> </msub> </msubsup> <mi>&amp;phi;</mi> <mi>d</mi> <mi>&amp;phi;</mi> <mo>=</mo> <mfrac> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </mfrac> <msubsup> <mi>&amp;phi;</mi> <mi>K</mi> <mn>2</mn> </msubsup> <mo>=</mo> <mfrac> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>-</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> <mrow> <mn>2</mn> <msub> <mi>r</mi> <mi>b</mi> </msub> </mrow> </mfrac> </mrow>
Similarly, arc BA and arc QA length are respectively
<mrow> <msub> <mi>L</mi> <mrow> <mi>B</mi> <mi>A</mi> </mrow> </msub> <mo>=</mo> <mfrac> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </mfrac> <msubsup> <mi>&amp;phi;</mi> <mi>B</mi> <mn>2</mn> </msubsup> </mrow>
<mrow> <msub> <mi>L</mi> <mrow> <mi>Q</mi> <mi>A</mi> </mrow> </msub> <mo>=</mo> <mfrac> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </mfrac> <msubsup> <mi>&amp;phi;</mi> <mi>Q</mi> <mn>2</mn> </msubsup> <mo>=</mo> <mfrac> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </mfrac> <msup> <mi>tan</mi> <mn>2</mn> </msup> <msub> <mi>&amp;alpha;</mi> <mn>0</mn> </msub> </mrow>
Therefore arc BK length is
<mrow> <msub> <mi>L</mi> <mrow> <mi>B</mi> <mi>K</mi> </mrow> </msub> <mo>&amp;ap;</mo> <mfrac> <msub> <mi>h</mi> <mi>max</mi> </msub> <mrow> <msubsup> <mi>mh</mi> <mi>a</mi> <mo>*</mo> </msubsup> </mrow> </mfrac> <msub> <mi>L</mi> <mrow> <mi>Q</mi> <mi>K</mi> </mrow> </msub> <mo>=</mo> <mfrac> <msub> <mi>h</mi> <mi>max</mi> </msub> <mrow> <msubsup> <mi>mh</mi> <mi>a</mi> <mo>*</mo> </msubsup> </mrow> </mfrac> <mrow> <mo>(</mo> <mfrac> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>-</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> <mrow> <mn>2</mn> <msub> <mi>r</mi> <mi>b</mi> </msub> </mrow> </mfrac> <mo>-</mo> <mfrac> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </mfrac> <msup> <mi>tan</mi> <mn>2</mn> </msup> <msub> <mi>&amp;alpha;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow>
I.e.
<mrow> <mfrac> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </mfrac> <msubsup> <mi>&amp;phi;</mi> <mi>B</mi> <mn>2</mn> </msubsup> <mo>=</mo> <mfrac> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>-</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> <mrow> <mn>2</mn> <msub> <mi>r</mi> <mi>b</mi> </msub> </mrow> </mfrac> <mo>-</mo> <mfrac> <msub> <mi>h</mi> <mi>max</mi> </msub> <mrow> <msubsup> <mi>mh</mi> <mi>a</mi> <mo>*</mo> </msubsup> </mrow> </mfrac> <mrow> <mo>(</mo> <mfrac> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>-</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> <mrow> <mn>2</mn> <msub> <mi>r</mi> <mi>b</mi> </msub> </mrow> </mfrac> <mo>-</mo> <mfrac> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </mfrac> <msup> <mi>tan</mi> <mn>2</mn> </msup> <msub> <mi>&amp;alpha;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow>
So roll angle φs of the point B in involuteBFor
<mrow> <msub> <mi>&amp;phi;</mi> <mi>B</mi> </msub> <mo>=</mo> <msqrt> <mrow> <mfrac> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>-</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mfrac> <mo>-</mo> <mfrac> <msub> <mi>h</mi> <mi>max</mi> </msub> <mrow> <msubsup> <mi>mh</mi> <mi>a</mi> <mo>*</mo> </msubsup> </mrow> </mfrac> <mrow> <mo>(</mo> <mfrac> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>-</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mfrac> <mo>-</mo> <msup> <mi>tan</mi> <mn>2</mn> </msup> <msub> <mi>&amp;alpha;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </msqrt> </mrow>
Coordinates of the point B in coordinate system XOY, which can be obtained, is
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>x</mi> <mi>B</mi> </msub> <mo>=</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>cos&amp;phi;</mi> <mi>B</mi> </msub> <mo>+</mo> <msub> <mi>&amp;phi;</mi> <mi>B</mi> </msub> <msub> <mi>sin&amp;phi;</mi> <mi>B</mi> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>y</mi> <mi>B</mi> </msub> <mo>=</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>sin&amp;phi;</mi> <mi>B</mi> </msub> <mo>-</mo> <msub> <mi>&amp;phi;</mi> <mi>B</mi> </msub> <msub> <mi>cos&amp;phi;</mi> <mi>B</mi> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced>
1-1-3, determine point B in coordinate system X1OY1In coordinate be
<mrow> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <msub> <mi>x</mi> <mrow> <mn>1</mn> <mi>B</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>y</mi> <mrow> <mn>1</mn> <mi>B</mi> </mrow> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>=</mo> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> <mtd> <mrow> <mo>-</mo> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> <mtd> <mrow> <mi>sin</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>cos&amp;phi;</mi> <mi>B</mi> </msub> <mo>+</mo> <msub> <mi>&amp;phi;</mi> <mi>B</mi> </msub> <msub> <mi>sin&amp;phi;</mi> <mi>B</mi> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>sin&amp;phi;</mi> <mi>B</mi> </msub> <mo>-</mo> <msub> <mi>&amp;phi;</mi> <mi>B</mi> </msub> <msub> <mi>cos&amp;phi;</mi> <mi>B</mi> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
A kind of 4. involute profile correction method of harmonic drive according to claim 3, it is characterised in that：The step Determine point E in coordinate system X in 1-11OY1In coordinate method it is as follows, must can be exhibited angle θ according to point K roll angleKFor
θK=φK-arctanφk
Therefore
βK=β0+θ0-θK
<mrow> <msub> <mi>&amp;beta;</mi> <mi>E</mi> </msub> <mo>=</mo> <msub> <mi>&amp;beta;</mi> <mi>K</mi> </msub> <mo>-</mo> <mfrac> <msub> <mi>&amp;Delta;</mi> <mi>max</mi> </msub> <msub> <mi>r</mi> <mi>a</mi> </msub> </mfrac> </mrow>
Wherein
So point E is in coordinate system X1OY1In coordinate be
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>x</mi> <mrow> <mn>1</mn> <mi>E</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>r</mi> <mi>a</mi> </msub> <msub> <mi>sin&amp;beta;</mi> <mi>E</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>y</mi> <mrow> <mn>1</mn> <mi>E</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>r</mi> <mi>a</mi> </msub> <msub> <mi>cos&amp;beta;</mi> <mi>E</mi> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced>
A kind of 5. involute profile correction method of harmonic drive according to claim 4, it is characterised in that：According to described Step 1-1-3 midpoint B are in coordinate system X1OY1In coordinate derivation can show that slope of the involute at point B is
<mrow> <msub> <mi>k</mi> <mrow> <mn>1</mn> <mi>B</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mo>+</mo> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> <msub> <mi>tan&amp;phi;</mi> <mi>B</mi> </msub> </mrow> <mrow> <mi>tan</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;beta;</mi> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> <mo>-</mo> <msub> <mi>tan&amp;phi;</mi> <mi>B</mi> </msub> </mrow> </mfrac> </mrow>
A kind of 6. involute profile correction method of harmonic drive according to claim 5, it is characterised in that：At tooth top The position of the E points determines that the maximum profiling quantity is relevant with the modulus m of the flexbile gear gear teeth, maximum correction of the flank shape by maximum profiling quantity The span of amount is (0.05m, 0.4m).
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