CN103714205A - Clock spring model simulating rotatable deformation of flexible body - Google Patents

Abstract

The invention provides a clock spring model simulating rotatable deformation of a flexible body. A plurality of clock springs are sequentially connected in series to form the clock spring model; in an interactive process, signals which are fed back as force touch information which is calculated by the clock spring model and responds to real-time deformation simulation of the flexible body under the action of torque are output, and the sum of the quantities of torsional deformation of all circles of the clock spring model is equivalent to deformation of the surface of the flexible body. The same method is used for calculating the quantities of the torsional deformation of all the circles of the clock spring model, calculation is simple, and the speed of torsional deformation calculation is increased; different types of flexible bodies can be simulated by adjusting the radiuses of rotatable mandrels of the clock springs, the thicknesses of the clock springs and the suspension section widths, and applicability is high; the clock spring model can be applied to the fields of virtual surgery simulation, remote operation robot control, telemedicine and the like.

Description

A kind of clockwork spring model of simulating the rotatable distortion of flexible body

Technical field

The invention belongs to computer simulation technique field, relate in particular to a kind of clockwork spring model of simulating the rotatable distortion of flexible body.

Background technology

Virtual operation emulation is the important application of virtual reality technology in modern operative training, mainly by visual, the moment of torsion deformation emulating stressed with modeling, human body soft tissue organ of medical data, formed, the true reappearance of surgical scene is provided for user on vision and sense of touch sense organ.Compare with traditional operation training, virtual operation can be simulated various physics and the physiological property of biological tissue, has greatly improved the effect of operative training; The experimental subjects that can recycle that simultaneously can also be different according to need to providing of doctor, greatly reduces the cost of surgeon's operative training.

It is the gordian technique of virtual operation emulation that the human body flexible bodily form becomes.At present, mainly contain the deformation models such as Mass-spring model, finite element model, boundary element model.That spring-particle modeling method has is simple, calculated amount is less, be easy to the advantages such as realization, but has the drawbacks such as topological analysis difficulty.Although finite element modeling method can carry out accurate and quantitative simulation to the deformation of object, relate to a large amount of numerous and diverse calculating, real-time, interactive poor performance.Though it is discrete that boundary element modeling method is carried out the border of modeling method, simplified calculating, aspect stability, there is certain deficiency.More than explanation, all there is the problems such as the comparatively numerous and diverse and simulation accuracy of calculating is not high in these conventional flexible bodies distortion Method of Physical Modeling, has then affected real-time and the validity calculated.Therefore, set up the accurately haptic modeling method of efficient virtual flexible body deformation simulation, become current fictitious force haptic-display system matter of utmost importance urgently to be resolved hurrily.

In view of the above problems, in virtual operation simulation process, the combination of vision and power sense of touch makes virtual emulation truer, credible, has proposed a kind of clockwork spring model of simulating the rotatable deformation simulation of flexible body.

Summary of the invention

Technical matters to be solved by this invention is to overcome the deficiencies in the prior art, and a kind of clockwork spring model of simulating the rotatable distortion of flexible body is provided.

For solving the problems of the technologies described above, the technical solution adopted in the present invention is:

A clockwork spring model of simulating the rotatable distortion of flexible body, comprises the steps:

Step 1, sets up rectangular coordinate system in space, determines the clockwork spring of taking up an official post flexible body surface meaning point place laying, and its process is as follows:

Step 1.1, sets up rectangular coordinate system in space,

Under given moment of torsion M effect, on flexible body surface, a clockwork spring is laid at O place, arbitrfary point, is being that r place arranges the rotatable mandrel points outside of clockwork spring A apart from arbitrfary point O ₀, r is the rotatable mandrel radius of clockwork spring, take arbitrfary point O as initial point, ray OA ₀place direction is X-axis positive dirction, sets up XYZ space coordinate system;

Step 1.2, arranges clockwork spring by circle,

Wherein, A outside i-1 circle clockwork spring _iX1point, suspension thickness is h _i=h ₁+ (i-1) i of d encloses clockwork spring, wherein h ₁be the thickness of the 1st circle clockwork spring, d is given constant and d>0, i circle clockwork spring with initial point O (0,0,0) for the center of circle, with

for the circle of radius, i=1,2,3 ... S, S is natural number,

Wide b, the cross sectional moment of inertia of being of every circle clockwork spring

elastic modulus E depends on the material of flexible body, and flexible body material is all identical;

Step 2, determines the moment of torsion that arbitrary circle clockwork spring consumes;

Set active line and the rotatable mandrel points outside of the clockwork spring A of given moment of torsion M ₀place circle tangent, and under given moment of torsion M effect, if P circle clockwork spring produces torsional deflection before total in flexible body, P circle clockwork spring is called distortion check loop; And P≤S, namely the number of turns of clockwork spring at least equals P;

According to clockwork spring characteristic, set:

On front P-1 circle clockwork spring, any point is under given moment of torsion M effect, and the maximum twist power that consumes is equal and be F ₀; On P circle clockwork spring, any point is under given moment of torsion M effect, the maximum twist power consuming

all equate, and be not more than F ₀;

Step 2.1, in P-1 circle clockwork spring, is total to a little the moment of torsion M consuming before determining on i circle _ifor:

$M_i=F_0\·R_i=F_0\·(r+\underset{i=1}{\overset{i}{\mathrm{\Σ}}}{h}_i)$

Step 2.2, determines that P circle clockwork spring is total to a little the moment of torsion consuming

for:

$M_P^{\′}=M-\underset{i=1}{\overset{P-1}{\mathrm{\Σ}}}{M}_i$

On P circle clockwork spring, any point is under given moment of torsion M effect, the maximum twist power of consumption for:

$F_P^{\′}=\frac{M_P^{\′}}{R_P}=\frac{M_P^{\′}}{r+\underset{i=1}{\overset{P}{\mathrm{\Σ}}}{h}_i}$

Step 3, calculates on i circle clockwork spring institute a little, at the moment of torsion M of common consumption _iunder effect, the torsion number of turns n of generation _i;

$n_i=\left\{\begin{array}{cc}\frac{M_i{l}_i}{2\mathrm{\π}{\mathrm{EI}}_i}& i=\mathrm{1,2,3},...,P-1\\ \frac{M_P^{\′}{l}_i}{2\mathrm{\π}{\mathrm{EI}}_i}& i=P\end{array}\right.$

Wherein, l _ibe the effective length of i circle clockwork spring, its expression formula is as follows:

$l_i=\left\{\begin{array}{cc}2\mathrm{\π}(R_1+R_2+\·\·\·+R_i)& i=\mathrm{1,2,3},...,P-1\\ 2\mathrm{\π}(R_1+R_2+\·\·\·+R_{P-1})+\frac{M_P^{\′}}{F_0}& i=P\end{array}.\right.$

Described a kind of clockwork spring model of simulating the rotatable distortion of flexible body, the time delay time that described front P circle clockwork spring produces torsional deflection total to be needed is not more than 1ms, also meets the requirement that refreshing frequency is not less than 1000Hz.

Described a kind of clockwork spring model of simulating the rotatable distortion of flexible body, in all clockwork springs, the time delay time that every circle produces torsional deflection to be needed forms Geometric Sequence, meets:

t _i=q ^i-1t ₁

Wherein, t _irepresent the time delay time that i circle clockwork spring produces torsional deflection needs, the common ratio that q is Geometric Sequence, t ₁be that the 1st circle clockwork spring produces the time delay time that torsional deflection needs, 1≤i≤P.

Described a kind of clockwork spring model of simulating the rotatable distortion of flexible body, described clockwork spring model, on arbitrary circle clockwork spring, the twisting resistance of any point consumption reaches after maximum twist power, and its next circle starts to produce torsional deflection.

The invention has the beneficial effects as follows: the present invention proposes a kind of clockwork spring model of simulating the rotatable distortion of flexible body, described clockwork spring model is composed in series successively by a plurality of clockwork springs, in reciprocal process, the signal that output feedback is reaction power tactile data of flexible body real-time deformation emulation under torsional interaction of adopting clockwork spring model and calculating, the stack that produces torsional deflection amount sum in this clockwork spring model on all circles is externally equivalent to the distortion on flexible body surface.The torsional deflection amount computing method of the every circle clockwork spring of clockwork spring model of the present invention are identical, calculate simply, accelerated the speed that torsional deflection is calculated; By regulating the rotatable mandrel radius of clockwork spring, the thickness of clockwork spring and suspension cross-sectional width etc., just can simulate dissimilar flexible body, and applicability is wide; Can be applicable to the fields such as virtual surgery emulation, teleoperated vehicle's control, tele-medicine.

Accompanying drawing explanation

Fig. 1 is clockwork spring model schematic diagram.

Fig. 2 is flexible body deformation simulation process flow diagram.

Fig. 3 is clockwork spring model building method process flow diagram.

Fig. 4 is clockwork spring model moment of torsion, the torsion number of turns and time delay time relationship schematic diagram.

Embodiment

A kind of clockwork spring model of simulating the rotatable distortion of flexible body the present invention being proposed below in conjunction with flow process shown in accompanying drawing is elaborated:

Clockwork spring model schematic diagram as shown in Figure 1.A clockwork spring model of simulating the rotatable distortion of flexible body, its concrete steps are as follows:

Step 1, sets up rectangular coordinate system in space, determines the clockwork spring of taking up an official post flexible body surface meaning point place laying, and its process is as follows:

Step 1.1, sets up rectangular coordinate system in space,

Under given moment of torsion M effect, on flexible body surface, a clockwork spring is laid at O place, arbitrfary point, is being that r place arranges the rotatable mandrel points outside of clockwork spring A apart from arbitrfary point O ₀, r is the rotatable mandrel radius of clockwork spring, take arbitrfary point O as initial point, ray OA ₀place direction is X-axis positive dirction, sets up XYZ space coordinate system;

Step 1.2, arranges clockwork spring by circle,

Wherein, A outside i-1 circle clockwork spring _i-1point, suspension thickness is h _i=h ₁+ (i-1) i of d encloses clockwork spring, wherein h ₁be the thickness of the 1st circle clockwork spring, d is given constant and d>0, i circle clockwork spring with initial point O (0,0,0) for the center of circle, with for the circle of radius, i=1,2,3 ... S, S is natural number,

Wide b, the cross sectional moment of inertia of being of every circle clockwork spring

elastic modulus E depends on the material of flexible body, and flexible body material is all identical;

Step 2, determines the moment of torsion that arbitrary circle clockwork spring consumes;

Suppose active line and the rotatable mandrel points outside of the clockwork spring A of given moment of torsion M ₀place circle tangent, and under given moment of torsion M effect, if P circle clockwork spring produces torsional deflection before total in flexible body, P circle clockwork spring is called distortion check loop; And P≤S, namely the number of turns of clockwork spring at least equals P;

According to clockwork spring characteristic, set:

On front P-1 circle clockwork spring, any point is under given moment of torsion M effect, and the maximum twist power that consumes is equal and be F ₀; On P circle clockwork spring, any point is under given moment of torsion M effect, the maximum twist power consuming

all equate, and be not more than F ₀;

Step 2.1, in P-1 circle clockwork spring, is total to a little the moment of torsion M consuming before determining on i circle _ifor:

$M_i=F_0\·R_i=F_0\·(r+\underset{i=1}{\overset{i}{\mathrm{\Σ}}}{h}_i)$

Step 2.2, determines that P circle clockwork spring is total to a little the moment of torsion consuming

for:

$M_P^{\′}=M-\underset{i=1}{\overset{P-1}{\mathrm{\Σ}}}{M}_i$

On P circle clockwork spring, any point is under given moment of torsion M effect, the maximum twist power of consumption

for:

$F_P^{\′}=\frac{M_P^{\′}}{R_P}=\frac{M_P^{\′}}{r+\underset{i=1}{\overset{P}{\mathrm{\Σ}}}{h}_i}$

Step 3, calculates on i circle clockwork spring institute a little, at the moment of torsion M of common consumption _iunder effect, the torsion number of turns n of generation _i;

$n_i=\left\{\begin{array}{cc}\frac{M_i{l}_i}{2\mathrm{\π}{\mathrm{EI}}_i}& i=\mathrm{1,2,3},...,P-1\\ \frac{M_P^{\′}{l}_i}{2\mathrm{\π}{\mathrm{EI}}_i}& i=P\end{array}\right.$

Wherein, l _ibe the effective length of i circle clockwork spring, its expression formula is as follows:

$l_i=\left\{\begin{array}{cc}2\mathrm{\π}(R_1+R_2+\·\·\·+R_i)& i=\mathrm{1,2,3},...,P-1\\ 2\mathrm{\π}(R_1+R_2+\·\·\·+R_{P-1})+\frac{M_P^{\′}}{F_0}& i=P\end{array}.\right.$

Described a kind of clockwork spring model of simulating the rotatable distortion of flexible body, the time delay time that described front P circle clockwork spring produces torsional deflection total to be needed is not more than 1ms, also meets the requirement that refreshing frequency is not less than 1000Hz, 1≤i≤P.

Described a kind of clockwork spring model of simulating the rotatable distortion of flexible body, in all clockwork springs, the time delay time that every circle produces torsional deflection to be needed forms Geometric Sequence, meets:

t _i=q ^i-1t ₁

Wherein, t _irepresent the time delay time that i circle clockwork spring produces torsional deflection needs, the common ratio that q is Geometric Sequence, t ₁be that the 1st circle clockwork spring produces the time delay time that torsional deflection needs.

Described a kind of clockwork spring model of simulating the rotatable distortion of flexible body, described clockwork spring model, on arbitrary circle clockwork spring, the twisting resistance of any point consumption reaches after maximum twist power, and its next circle starts to produce torsional deflection.

Virtual hand and Virtual Cardiac Mode take below as example, enumerate the embodiment of technical solution of the present invention.

In this example, all virtual hand and Virtual Cardiac Mode all directly adopt the OBJ form of deriving from 3DS MAX 2013 softwares, with 1558 particles, 3114 virtual hand and 3910 particles that triangle gridding forms, the Virtual Cardiac Mode of 7814 triangle gridding formations is that example is carried out deformation simulation, and in experimentation, model obtains and revise very convenient; Operating system is Windows 2000, take 3DS MAX 2013, OpenGL shape library as basis, on Microsoft Visual C++2012 Software Development Platform, carries out emulation.

When virtual hand being detected and collide on virtual heart surface any point, in given counterclockwise moment of torsion M=10 * 10 ^-3under Nm effect, the inner clockwork spring model of filling of regional area that virtual hand and virtual heart are mutual, in reciprocal process, output is fed back to the signal that adopts reaction power tactile data of virtual heart real-time deformation emulation under torsional interaction that clockwork spring model calculates, as shown in Figure 2;

Under given counterclockwise moment of torsion M effect, when virtual hand collides on virtual heart surface arbitrfary point O, on virtual heart surface, a clockwork spring is laid at O place, arbitrfary point, apart from arbitrfary point O, is being r=1 * 10 ^-3m place arranges the rotatable mandrel points outside of clockwork spring A ₀, r is the rotatable mandrel radius of clockwork spring, take arbitrfary point O as initial point, ray OA ₀place direction is X-axis positive dirction, sets up XYZ space coordinate system;

Set gradually each circle of clockwork spring, as shown in figures 1 and 3; Wherein, A outside i-1 circle clockwork spring _i-1point, suspension thickness is h _i=h ₁+ (i-1) i of the clockwork spring of d encloses, h ₁=0.4 * 10 ^-3m, d=0.1 * 10 ^-3m, i circle clockwork spring with initial point O (0,0,0) for the center of circle, with

for the circle of radius, i=1,2,3 ... S, S is natural number;

With O (0,0,0) for the center of circle, with

R ₁=r+h ₁=1 * 10 ^-3+ 0.4 * 10 ^-3=1.4 * 10 ^-3m is the circle of radius, forms the 1st circle clockwork spring;

R ₂=r+h ₁+ h ₂=r+2h ₁+ (2-1) d=1 * 10 ^-3+ 2 * 0.4 * 10 ^-3+ 1 * 0.1 * 10 ^-3=1.9 * 10 ^-3m is the circle of radius, forms the 2nd circle clockwork spring;

R ₃=r+h ₁+ h ₂+ h ₃=r+3h ₁+ 3d=1 * 10 ^-3+ 3 * 0.4 * 10 ^-3+ 3 * 0.1 * 10 ^-3=2.5 * 10 ^-3m is the circle of radius, forms the 3rd circle clockwork spring;

R ₄=r+h ₁+ h ₂+ h ₃+ h ₄=r+4h ₁+ 6d=1 * 10 ^-3+ 4 * 0.4 * 10 ^-3+ 6 * 0.1 * 10 ^-3=3.2 * 10 ^-3m is the circle of radius, forms the 4th circle clockwork spring;

R ₅=r+h ₁+ h ₂+ h ₃+ h ₄+ h ₅=r+5h ₁+ 10d=1 * 10 ^-3+ 5 * 0.4 * 10 ^-3+ 10 * 0.1 * 10 ^-3=4 * 10 ^-3m is the circle of radius, forms the 5th circle clockwork spring;

The radius R of i circle clockwork spring _iform

ordered series of numbers;

Suppose that the wide of every circle clockwork spring is b=6 * 10 ^-3m, elastic modulus E=3.09 * 10 ⁷pa depends on the material of flexible body, all identical;

The pilot process calculating, the equal round off method of last data retain after radix point 3.

Suppose given counterclockwise moment of torsion M=10 * 10 ^-3the rotatable mandrel points outside of the active line of Nm and clockwork spring A ₀place circle tangent, and under given moment of torsion M effect, if P circle clockwork spring produces torsional deflection before total in flexible body, P circle clockwork spring is called distortion check loop;

According to clockwork spring property settings: as shown in Figure 4; On front P-1 circle clockwork spring, any point is under given counterclockwise moment of torsion M effect, and the maximum twist power consuming equates and is F ₀=0.989N, on P circle clockwork spring any point under given counterclockwise moment of torsion M effect, the maximum twist power consuming

all equate, and be not more than F ₀=0.989N;

The time delay time that front i circle clockwork spring produces torsional deflection total to be needed is not more than 1ms, 1≤i≤P; And in each circle of all clockwork springs, the time delay time formation that every circle produces torsional deflection to be needed produces with the 1st circle clockwork spring the time delay time t that torsional deflection needs ₁=10 ^-5the Geometric Sequence that s is first term, the q=1.2 of take is common ratio; On arbitrary circle clockwork spring, the twisting resistance of any point consumption reaches after maximum twist power, and its next circle starts to produce torsional deflection.

Suppose that haptic feedback refreshing frequency is 1200Hz, the inverse of haptic feedback refreshing frequency $T=\frac{1}{1200}s;$

If any point is under given counterclockwise moment of torsion M effect on the 1st circle clockwork spring, the maximum twist power consuming equates also all to arrive F ₀, on the 1st circle clockwork spring, be total to a little the moment of torsion M consuming ₁for:

M ₁=F ₀R ₁=0.989×1.4×10 ^-3=1.385×10 ^-3N·m,

M ₁<M,

The effective length l of the 1st circle clockwork spring ₁=2 π R ₁,

The cross sectional moment of inertia of the 1st circle clockwork spring $I_1=\frac{b{h}_1^3}{12}=\frac{6\&times;{10}^{-3}\&times;{(0.4\&times;{10}^{-3})}^3}{12}=32\&times;{10}^{-15}{m}^4,$

The 1st circle clockwork spring on institute a little, at the moment of torsion M of common consumption ₁under effect, with the torsion number of turns n of its generation ₁between meet:

$n_1=\frac{M_1{l}_1}{2\mathrm{\&pi;}{\mathrm{EI}}_1}=\frac{1.385\&times;{10}^{-3}\&times;2\mathrm{\&pi;}\&times;1.4\&times;{10}^{-3}}{2\mathrm{\&pi;}\&times;3.09\&times;{10}^7\&times;32\&times;{10}^{-15}}=1.961,$

The 1st circle clockwork spring produces the time delay time T that torsional deflection needs ₁=t ₁=10 ^-5s<T;

Therefore, on the 1st circle clockwork spring, be total to a little the moment of torsion M consuming ₁<M, and the 1st circle clockwork spring produces the time delay time T that torsional deflection total needs ₁=10 ^-5s<T, meets the requirement of refreshing frequency; Only has the maximum twist power F that reaches consumption when the 1st twisting resistance that encloses any point consumption on clockwork spring ₀after, the 2nd circle clockwork spring just starts to produce torsional deflection.

If any point is under given counterclockwise moment of torsion M effect on the 2nd circle clockwork spring, the maximum twist power consuming equates also all to arrive F ₀, on the 2nd circle clockwork spring, be total to a little the moment of torsion M consuming ₂for:

M ₂=F ₀R ₂=0.989×1.9×10 ^-3=1.879×10 ^-3N·m，

On front 2 circle clockwork springs, being total to a little the moment of torsion sum consuming is: M ₁+ M ₂=1.385 * 10 ^-3+ 1.879 * 10 ^-3=3.264 * 10 ^-3nm<M=10 * 10 ^-3nm,

The effective length l of the 2nd circle clockwork spring ₂=2 π (R ₁+ R ₂),

The cross sectional moment of inertia of the 2nd circle clockwork spring:

$I_2=\frac{b{h}_2^3}{12}=\frac{b{(h_1+d)}^3}{12}=\frac{6\&times;{10}^{-3}\&times;{(0.4\&times;{10}^{-3}+0.1\&times;{10}^{-3})}^3}{12}=62.5\&times;{10}^{-15}{m}^4,$

The 2nd circle clockwork spring on institute a little, at the moment of torsion M of common consumption ₂under effect, with the torsion number of turns n of its generation ₂between meet:

$n_2=\frac{M_2{l}_2}{2\mathrm{\&pi;}{\mathrm{EI}}_2}=\frac{1.879\&times;{10}^{-3}\&times;2\mathrm{\&pi;}\&times;(1.4\&times;{10}^{-3}+1.9\&times;{10}^{-3})}{2\mathrm{\&pi;}\&times;3.09\&times;{10}^7\&times;62.5\&times;{10}^{-15}}=3.211,$

Front 2 circle clockwork springs produce the time delay time T that torsional deflection needs ₂=t ₁+ t ₂=(1+q) t ₁=(1+1.2) * 10 ^-5=2.2 * 10 ^-5s<T, T is the inverse of haptic feedback refreshing frequency here,

Therefore, on front 2 circle clockwork springs, be total to a little the moment of torsion M consuming ₁+ M ₂<M, and front 2 circle clockwork springs produce the time delay time T that torsional deflection total needs ₂=2.2 * 10 ^-5s<T, meets the requirement of refreshing frequency; Only has the maximum twist power F that reaches consumption when the 2nd twisting resistance that encloses any point consumption on clockwork spring ₀after, the 3rd circle clockwork spring just starts to produce torsional deflection.

If any point is under given counterclockwise moment of torsion M effect on the 3rd circle clockwork spring, the maximum twist power consuming equates also all to arrive F ₀, on the 3rd circle clockwork spring, be total to a little the moment of torsion M consuming ₃for:

M ₃=F ₀R ₃=0.989×2.5×10 ^-3=2.473×10 ^-3N·m，

On front 3 circle clockwork springs, being total to a little the moment of torsion sum consuming is:

M ₁+M ₂+M ₃=1.385×10 ^-3+1.879×10 ^-3+2.473×10 ^-3=5.737×10 ^-3N·m<M=10×10 ^-3N·m

The effective length l of the 3rd circle clockwork spring ₃=2 π (R ₁+ R ₂+ R ₃),

The cross sectional moment of inertia of the 3rd circle clockwork spring:

$I_3=\frac{b{h}_3^3}{12}=\frac{b{(h_1+2d)}^3}{12}=\frac{6\&times;{10}^{-3}\&times;{(0.4\&times;{10}^{-3}+2\&times;0.1\&times;{10}^{-3})}^3}{12}=108\&times;{10}^{-15}{m}^4,$

The 3rd circle clockwork spring on institute a little, at the moment of torsion M of common consumption ₃under effect, with the torsion number of turns n of its generation ₃between meet:

$n_3=\frac{M_3{l}_3}{2\mathrm{\&pi;}{\mathrm{EI}}_3}=\frac{2.473\&times;{10}^{-3}\&times;2\mathrm{\&pi;}\&times;(1.4\&times;{10}^{-3}+1.9\&times;{10}^{-3}+2.5\&times;{10}^{-3})}{2\mathrm{\&pi;}\&times;3.09\&times;{10}^7\&times;108\&times;{10}^{-15}}=4.298,$

Front 3 circle clockwork springs produce the time delay time that torsional deflection needs:

T ₃=t ₁+t ₂+t ₃=(1+q+q ²)t ₁=(1+1.2+1.2 ²)×10 ^-5=3.64×10 ^-5s<T，

Here T is the inverse of haptic feedback refreshing frequency,

Therefore, on front 3 circle clockwork springs, be total to a little the moment of torsion M consuming ₁+ M ₂+ M ₃<M, and front 3 circle clockwork springs produce the time delay time T that torsional deflection total needs ₃=3.64 * 10 ^-5s<T, meets the requirement of refreshing frequency; Only has the maximum twist power F that reaches consumption when the 3rd twisting resistance that encloses any point consumption on clockwork spring ₀after, the 4th circle clockwork spring just starts to produce torsional deflection.

If any point is under given counterclockwise moment of torsion M effect on the 4th circle clockwork spring, the maximum twist power consuming equates also all to arrive F ₀, on the 4th circle clockwork spring, be total to a little the moment of torsion M consuming ₄for:

M ₄=F ₀R ₄=0.989×3.2×10 ^-3=3.165×10 ^-3N·m，

On front 4 circle clockwork springs, being total to a little the moment of torsion sum consuming is:

M ₁+M ₂+M ₃+M ₄=1.385×10 ^-3+1.879×10 ^-3+2.473×10 ^-3+3.165×10 ^-3，

=8.902×10 ^-3N·m<M=10×10 ^-3N·m

The effective length l of the 4th circle clockwork spring ₄=2 π (R ₁+ R ₂+ R ₃+ R ₄),

The cross sectional moment of inertia of the 4th circle clockwork spring:

$I_4=\frac{b{h}_4^3}{12}=\frac{b{(h_1+3d)}^3}{12}=\frac{6\&times;{10}^{-3}\&times;{(0.4\&times;{10}^{-3}+3\&times;0.1\&times;{10}^{-3})}^3}{12}=171.5\&times;{10}^{-15}{m}^4,$

The 4th circle clockwork spring on institute a little, at the moment of torsion M of common consumption ₄under effect, with the torsion number of turns n of its generation ₄between meet:

$n_4=\frac{M_4{l}_4}{2\mathrm{\&pi;}{\mathrm{EI}}_4}=\frac{3.165\&times;{10}^{-3}\&times;2\mathrm{\&pi;}\&times;(1.4\&times;{10}^{-3}+1.9\&times;{10}^{-3}+2.5\&times;{10}^{-3}+3.2\&times;{10}^{-3})}{2\mathrm{\&pi;}\&times;3.09\&times;{10}^7\&times;171.5\&times;{10}^{-15}}=5.375$

Front 4 circle clockwork springs produce the time delay time that torsional deflection needs:

T ₄=t ₁+t ₂+t ₃+t ₄=(1+q+q ²+q ³)t ₁=5.368×10 ^-5s<T，

Here T is the inverse of haptic feedback refreshing frequency,

Therefore, on front 4 circle clockwork springs, be total to a little the moment of torsion M consuming ₁+ M ₂+ M ₃+ M ₄<M, and front 4 circle clockwork springs generation torsional deflections amount to T between the time delay needing ₄=5.368 * 10 ^-5s<T, meets the requirement of refreshing frequency; Only has the maximum twist power F that reaches consumption when the 4th twisting resistance that encloses any point consumption on clockwork spring ₀after, the 5th circle clockwork spring just starts to produce torsional deflection.

If any point is under given counterclockwise moment of torsion M effect on the 5th circle clockwork spring, the maximum twist power consuming equates also all to arrive F ₀, on the 5th circle clockwork spring, be total to a little the moment of torsion M consuming ₅for:

M ₅=F ₀R ₅=0.989×4×10 ^-3=3.956×10 ^-3N·m，

On front 5 circle clockwork springs, being total to a little the moment of torsion sum consuming is:

M ₁+M ₂+M ₃+M ₄+M ₅=（1.385+1.879+2.473+3.165+3.956）×10 ^-3

=12.858×10 ^-3N·m>M=10×10 ^-3N·m

Therefore, on front 5 circle clockwork springs, be total to a little the moment of torsion M consuming ₁+ M ₂+ M ₃+ M ₄+ M ₅>M, is total to a little the moment of torsion sum consuming on front 5 circle clockwork springs and is not less than given counterclockwise moment of torsion M, and the 5th circle clockwork spring, for distortion check loop, does not need to judge whether to meet the requirement of refreshing frequency again;

The 5th circle clockwork spring is out of shape the moment of torsion that is total to a little consumption on check loop

for:

$M_5^{\&prime;}=M-(M_1+M_2+M_3+M_4)=10\&times;{10}^{-3}-8.902\&times;{10}^{-3}=1.098\&times;{10}^{-3}N\&CenterDot;m,$

Distortion check loop also on the 5th circle clockwork spring any point under given counterclockwise moment of torsion M effect, the maximum twist power of consumption

for:

$F_5^{\&prime;}=\frac{M_5^{\&prime;}}{R_5}=\frac{1.098\&times;{10}^{-3}}{4\&times;{10}^{-3}}=0.275N<F_0=0.989N,$

According to clockwork spring property settings: on front 4 circle clockwork springs, any point is under given counterclockwise moment of torsion M effect, and the maximum twist power of consumption equates to be also all assumed to F ₀=0.989N, distortion check loop also on the 5th circle clockwork spring any point under given counterclockwise moment of torsion M effect, the maximum twist power of consumption

all equate to be also all not more than F ₀=0.989N;

The effective length of the 5th circle clockwork spring $l_5=2\mathrm{\&pi;}(R_1+R_2+R_3+R_4)+\frac{M_5^{\&prime;}}{F_0},$

The cross sectional moment of inertia of the 5th circle clockwork spring:

$I_5=\frac{b{h}_5^3}{12}=\frac{b{(h_1+4d)}^3}{12}=\frac{6\&times;{10}^{-3}\&times;{(0.4\&times;{10}^{-3}+4\&times;0.1\&times;{10}^{-3})}^3}{12}=256\&times;{10}^{-15}{m}^4,$

The 5th circle clockwork spring is out of shape on check loop institute a little, at the moment of torsion M of common consumption ₅under effect, with the torsion number of turns n of its generation ₅between meet:

$n_5=\frac{M_5^{\&prime;}{l}_5}{2\mathrm{\&pi;}{\mathrm{EI}}_5}$

$=\frac{1.098\&times;{10}^{-3}\&times;[2\mathrm{\&pi;}\&times;(1.4\&times;{10}^{-3}+1.9\&times;{10}^{-3}+2.5\&times;{10}^{-3}+3.2\&times;{10}^{-3})+\frac{1.098\&times;{10}^{-3}}{0.989}]}{2\mathrm{\&pi;}\&times;3.09\&times;{10}^7\&times;256\&times;{10}^{-15}}=1.274$

Front 5 circle clockwork springs are in given counterclockwise moment of torsion M=10 * 10 ^-3under Nm effect, the raw torsion of common property several n that turn-take are:

n=n ₁+n ₂+n ₃+n ₄+n ₅=1.961+3.211+4.298+5.375+1.274=16.119。

Attention: adopting the clockwork spring model of the rotatable distortion of flexible body to calculate under given counterclockwise torsional interaction, in the process of flexible body real-time deformation emulation, if r, d, h ₁, these parameters of b choose excessive, in the modeling method of the clockwork spring model of the rotatable distortion of flexible body, the distortion check loop numerical value of clockwork spring is just few, calculated amount is little, real-time is good, but deformation simulation poor effect; If r, d, h ₁, these parameters of b choose too small, in the modeling method of the clockwork spring model of the rotatable distortion of flexible body, the distortion check loop numerical value of clockwork spring is just large, calculated amount is large, real-time is not good, but deformation simulation effect is better; In addition t is being set ₁and t _ibetween be related to time, should be taken into account the hardware configuration of program operation computer-chronograph itself, therefore in the process of the whole program of debugging, compromise and select these parameters, constantly repeatedly debug, thereby make deformation effect more true to nature.

For verifying implementation result of the present invention, operator is by the deformation simulation that the handle of PHANTOM OMNI hand controller end touches, perception and control virtual hand are rotated virtual heart, and by the power tactile data Real-time Feedback producing in reciprocal process to operator.Experimental result shows: this model is simply effective, and deformation effect is true to nature, and image is smooth, and power tactile sensation is steady, and operator to the perception of virtual environment and alternately accurately and reliably, can meet the requirement of virtual operation simulation interactive system.

Claims (4)

1. a clockwork spring model of simulating the rotatable distortion of flexible body, is characterized in that, comprises the steps:

Step 1, sets up rectangular coordinate system in space, determines the clockwork spring of taking up an official post flexible body surface meaning point place laying, and its process is as follows:

Step 1.1, sets up rectangular coordinate system in space,

Under given moment of torsion M effect, on flexible body surface, a clockwork spring is laid at O place, arbitrfary point, is being that r place arranges the rotatable mandrel points outside of clockwork spring A apart from arbitrfary point O ₀, r is the rotatable mandrel radius of clockwork spring, take arbitrfary point O as initial point, ray OA ₀place direction is X-axis positive dirction, sets up XYZ space coordinate system;

Step 1.2, arranges clockwork spring by circle,

Wherein, A outside i-1 circle clockwork spring _i-1point, suspension thickness is h _i=h ₁+ (i-1) i of d encloses clockwork spring, wherein h ₁be the thickness of the 1st circle clockwork spring, d is given constant and d>0, i circle clockwork spring be with initial point O (0,0,0) for the center of circle, with

for the circle of radius, i=1,2,3 ... S, S is natural number,

Wide b, the cross sectional moment of inertia of being of every circle clockwork spring elastic modulus E depends on the material of flexible body, and flexible body material is all identical;

Step 2, determines the moment of torsion that arbitrary circle clockwork spring consumes;

Set active line and the rotatable mandrel points outside of the clockwork spring A of given moment of torsion M ₀place circle tangent, and under given moment of torsion M effect, if P circle clockwork spring produces torsional deflection before total in flexible body, P circle clockwork spring is called distortion check loop; And P≤S, namely the number of turns of clockwork spring at least equals P;

According to clockwork spring characteristic, set:

On front P-1 circle clockwork spring, any point is under given moment of torsion M effect, and the maximum twist power that consumes is equal and be F ₀; On P circle clockwork spring, any point is under given moment of torsion M effect, the maximum twist power consuming

all equate, and be not more than F ₀;

Step 2.1, in P-1 circle clockwork spring, is total to a little the moment of torsion M consuming before determining on i circle _ifor:

$M_i=F_0\&CenterDot;R_i=F_0\&CenterDot;(r+\underset{i=1}{\overset{i}{\mathrm{\&Sigma;}}}{h}_i)$

Step 2.2, determines that P circle clockwork spring is total to a little the moment of torsion consuming

for:

$M_P^{\&prime;}=M-\underset{i=1}{\overset{P-1}{\mathrm{\&Sigma;}}}{M}_i$

On P circle clockwork spring, any point is under given moment of torsion M effect, the maximum twist power of consumption

for:

$F_P^{\&prime;}=\frac{M_P^{\&prime;}}{R_P}=\frac{M_P^{\&prime;}}{r+\underset{i=1}{\overset{P}{\mathrm{\&Sigma;}}}{h}_i}$

Step 3, calculates on i circle clockwork spring institute a little, at the moment of torsion M of common consumption _iunder effect, the torsion number of turns n of generation _i;

$n_i=\left\{\begin{array}{cc}\frac{M_i{l}_i}{2\mathrm{\&pi;}{\mathrm{EI}}_i}& i=\mathrm{1,2,3},...,P-1\\ \frac{M_P^{\&prime;}{l}_i}{2\mathrm{\&pi;}{\mathrm{EI}}_i}& i=P\end{array}\right.$

Wherein, l _ibe the effective length of i circle clockwork spring, its expression formula is as follows:

$l_i=\left[\begin{array}{cc}2\mathrm{\&pi;}(R_1+R_2+\&CenterDot;\&CenterDot;\&CenterDot;+R_i)& i=\mathrm{1,2,3},...,P-1\\ 2\mathrm{\&pi;}(R_1+R_2+\&CenterDot;\&CenterDot;\&CenterDot;+R_{P-1})+\frac{M_P^{\&prime;}}{F_0}& i=P\end{array},.\right]$

2. a kind of clockwork spring model of simulating the rotatable distortion of flexible body according to claim 1, is characterized in that, the time delay time that described front P circle clockwork spring produces torsional deflection total to be needed is not more than 1ms.

3. a kind of clockwork spring model of simulating the rotatable distortion of flexible body according to claim 1, is characterized in that, in all clockwork springs, the time delay time that every circle produces torsional deflection to be needed forms Geometric Sequence, meets:

t _i=q ^i-1t ₁

Wherein, t _irepresent the time delay time that i circle clockwork spring produces torsional deflection needs, the common ratio that q is Geometric Sequence, t ₁be that the 1st circle clockwork spring produces the time delay time that torsional deflection needs, 1≤i≤P.

4. according to a kind of clockwork spring model of simulating the rotatable distortion of flexible body described in claim 1 or 2 or 3, it is characterized in that, described clockwork spring model, on arbitrary circle clockwork spring, the twisting resistance of any point consumption reaches after maximum twist power, and its next circle starts to produce torsional deflection.

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