CN103714205B - A kind of clockwork spring model simulating the rotatable deformation of flexible body - Google Patents

Abstract

The present invention proposes a kind of clockwork spring model simulating the rotatable deformation of flexible body, described clockwork spring model is sequentially connected in series by multiple clockwork springs and forms, in interaction, output feedback is the reaction using clockwork spring model to calculate signal of the power tactile data of flexible body real-time deformation emulation under moment of torsion effect, and the superposition producing torsional deflection amount sum in this clockwork spring model on all circles is externally equivalent to the deformation on flexible body surface.The torsional deflection amount computational methods that clockwork spring model of the present invention often encloses clockwork spring are identical, calculate simple, accelerate the speed that torsional deflection calculates；By regulation clockwork spring rotatable mandrel radius, the thickness of clockwork spring and suspension cross-sectional width, so that it may simulating different types of flexible body, the suitability is wide；Can be applicable to the fields such as virtual surgery emulation, teleoperated vehicle's control, tele-medicine.

Description

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

Technical field

The invention belongs to computer simulation technique field, particularly relate to a kind of clockwork spring bullet simulating the rotatable deformation of flexible body Spring model.

Background technology

Virtual teach-in is virtual reality technology important application in modem surgical is trained, mainly can by medical data Constitute with modeling, human body soft tissue organ stress, moment of torsion deformation emulating depending on changing, vision with sense of touch sense organ provide the user The true reappearance of surgical scene.Compared with traditional operation training, virtual operation can simulate various physics and the life of biological tissue Reason characteristic, significantly improves the effect of operative training；Simultaneously can also according to doctor need provide different can be repeatedly The experimental subject utilized, greatly reduces the cost of surgeon's operative training.

The change of the human body flexible bodily form is the key technology of virtual teach-in.At present, mainly there is mass spring model, limited The deformation model such as meta-model, boundary element model.Spring-mass modeling method have simple, amount of calculation is less, be easily achieved Etc. advantage, but there is the drawbacks such as topological analysis difficulty.Although the deformation of object can be carried out accurate and fixed by finite element modeling method The simulation of amount, but relate to the most numerous and diverse calculating, real-time, interactive poor performance.Though the border that boundary element modeling method is to modeling method Carry out discrete, simplify calculating, but in terms of stability, there is certain deficiency.Described above, these conventional flexible bodies All there is the problems such as the most numerous and diverse and simulation accuracy of calculating is the highest in deformation Method of Physical Modeling, then have impact on the real-time of calculating And effectiveness.Therefore, the haptic modeling method of accurate efficient virtual flexible body deformation simulation is set up, it has also become current virtual The matter of utmost importance that power haptic-display system is urgently to be resolved hurrily.

In view of the above problems, during virtual teach-in, the combination of vision and power sense of touch makes virtual emulation Truer, credible, it is proposed that a kind of clockwork spring model simulating the rotatable deformation simulation of flexible body.

Summary of the invention

The technical problem to be solved is to overcome the deficiencies in the prior art, it is provided that a kind of simulation flexible body can revolve Change the clockwork spring model of shape.

For solving above-mentioned technical problem, the technical solution adopted in the present invention is:

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

Step 1, sets up rectangular coordinate system in space, determines to take up an official post the clockwork spring laid at meaning point in flexible body surface, its mistake Journey is as follows:

Step 1.1, sets up rectangular coordinate system in space,

Under given moment of torsion M effect, at the arbitrfary point O of flexible body surface, lay a clockwork spring, at distance arbitrfary point O be One clockwork spring rotatable mandrel points outside A is set at r₀, i.e. r is clockwork spring rotatable mandrel radius, is former with arbitrfary point O Point, ray OA₀Direction, place is X-axis positive direction, sets up XYZ space coordinate system；

Step 1.2, arranges clockwork spring by circle,

Wherein, A outside the i-th-1 circle clockwork spring_iX1Point, suspension thickness is h_i=h₁The i-th circle clockwork spring of+(i-1) d, Wherein h₁Being the thickness of the 1st circle clockwork spring, d is given constant and d > 0, the i.e. i-th circle clockwork spring is with initial point O (0,0,0) For the center of circle, withFor the circle of radius, i=1,2,3 ... S, S are natural number,

The width often enclosing clockwork spring is b, cross sectional moment of inertiaElastic modulus E depends on the material of flexible body, And flexible body material is the most identical；

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

Set position and clockwork spring rotatable mandrel points outside A of given moment of torsion M₀The circle at place, place is tangent, and Under given moment of torsion M effect, if P circle clockwork spring produces torsional deflection before total in flexible body, then P circle clockwork spring is referred to as Deformation check loop；And P≤S, namely the number of turns of clockwork spring is at least equal to 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 consumed is equal and equal For F₀；On P circle clockwork spring, any point is under given moment of torsion M effect, the maximum twist power consumedThe most equal, and not More than F₀；

Step 2.1, determines in front P-1 circle clockwork spring, the i-th circle is total to a little the moment of torsion M consumed_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 consumedFor:

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

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

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

Step 3, on calculating the i-th circle clockwork spring, institute is a little, at the moment of torsion M of common consumption_iUnder effect, the torsion number of turns of generation n_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_iBeing the effective length of the i-th 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 simulating the rotatable deformation of flexible body, described front P circle clockwork spring produces to be turned round Change shape to amount to the decay time needed and be not more than 1ms, namely meet refreshing frequency and be not less than the requirement of 1000Hz.

Described a kind of clockwork spring model simulating the rotatable deformation of flexible body, in all clockwork springs, often encloses generation The decay time that torsional deflection needs constitutes Geometric Sequence, the most satisfied:

t_i=q^i-1t₁

Wherein, t_iRepresenting that the i-th circle clockwork spring produces the decay time that torsional deflection needs, q is the common ratio of Geometric Sequence, t₁It is that the 1st circle clockwork spring produces the decay time that torsional deflection needs, 1≤i≤P.

Described a kind of clockwork spring model simulating the rotatable deformation of flexible body, described clockwork spring model, arbitrary After on circle clockwork spring, the twisting resistance of any point consumption reaches maximum twist power, 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 mould simulating the rotatable deformation of flexible body Type, described clockwork spring model is sequentially connected in series by multiple clockwork springs and forms, and in interaction, output feedback is for using clockwork spring The reaction that spring model calculates is the signal of the power tactile data of flexible body real-time deformation emulation, this clockwork spring under moment of torsion effect The superposition producing torsional deflection amount sum in spring model on all circles is externally equivalent to the deformation on flexible body surface.The present invention sends out The torsional deflection amount computational methods that bar spring model often encloses clockwork spring are identical, calculate simple, accelerate what torsional deflection calculated Speed；By regulation clockwork spring rotatable mandrel radius, the thickness of clockwork spring and suspension cross-sectional width etc., so that it may simulation is not With the flexible body of type, the suitability is wide；Can be applicable to virtual surgery emulation, teleoperated vehicle controls, tele-medicine In field.

Accompanying drawing explanation

Fig. 1 is clockwork spring model schematic.

Fig. 2 is flexible body deformation simulation flow chart.

Fig. 3 is clockwork spring model building method flow chart.

Fig. 4 is clockwork spring model moment of torsion, reverses the number of turns and decay time relation schematic diagram.

Detailed description of the invention

A kind of clockwork spring simulating the rotatable deformation of flexible body that the present invention is proposed by the most shown flow process Model is described in detail:

Clockwork spring model schematic as shown in Figure 1.A kind of clockwork spring model simulating the rotatable deformation of flexible body, It specifically comprises the following steps that

Step 1, sets up rectangular coordinate system in space, determines to take up an official post the clockwork spring laid at meaning point in flexible body surface, its mistake Journey is as follows:

Step 1.1, sets up rectangular coordinate system in space,

Under given moment of torsion M effect, at the arbitrfary point O of flexible body surface, lay a clockwork spring, at distance arbitrfary point O be One clockwork spring rotatable mandrel points outside A is set at r₀, i.e. r is clockwork spring rotatable mandrel radius, is former with arbitrfary point O Point, ray OA₀Direction, place is X-axis positive direction, sets up XYZ space coordinate system；

Step 1.2, arranges clockwork spring by circle,

Wherein, A outside the i-th-1 circle clockwork spring_i-1Point, suspension thickness is h_i=h₁The i-th circle clockwork spring of+(i-1) d, Wherein h₁Being the thickness of the 1st circle clockwork spring, d is given constant and d > 0, the i.e. i-th circle clockwork spring is with initial point O (0,0,0) For the center of circle, withFor the circle of radius, i=1,2,3 ... S, S are natural number,

The width often enclosing clockwork spring is b, cross sectional moment of inertiaElastic modulus E depends on the material of flexible body, And flexible body material is the most identical；

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

Assuming that the position of given moment of torsion M and clockwork spring rotatable mandrel points outside A₀The circle at place, place is tangent, and Under given moment of torsion M effect, if P circle clockwork spring produces torsional deflection before total in flexible body, then P circle clockwork spring is referred to as Deformation check loop；And P≤S, namely the number of turns of clockwork spring is at least equal to 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 consumed is equal and equal For F₀；On P circle clockwork spring, any point is under given moment of torsion M effect, the maximum twist power consumedThe most equal, and not More than F₀；

Step 2.1, determines in front P-1 circle clockwork spring, the i-th circle is total to a little the moment of torsion M consumed_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 consumedFor:

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

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

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

Step 3, on calculating the i-th circle clockwork spring, institute is a little, at the moment of torsion M of common consumption_iUnder effect, the torsion number of turns of generation n_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_iBeing the effective length of the i-th 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 simulating the rotatable deformation of flexible body, described front P circle clockwork spring produces to be turned round Change shape to amount to the decay time needed and be not more than 1ms, namely meet refreshing frequency and be not less than the requirement of 1000Hz, 1≤i≤P.

Described a kind of clockwork spring model simulating the rotatable deformation of flexible body, in all clockwork springs, often encloses generation The decay time that torsional deflection needs constitutes Geometric Sequence, the most satisfied:

t_i=q^i-1t₁

Wherein, t_iRepresenting that the i-th circle clockwork spring produces the decay time that torsional deflection needs, q is the common ratio of Geometric Sequence, t₁It is that the 1st circle clockwork spring produces the decay time that torsional deflection needs.

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

Below as a example by virtual hand and Virtual Cardiac Mode, enumerate the detailed description of the invention of technical solution of the present invention.

In this example, all virtual hand and Virtual Cardiac Mode the most directly use derivation from 3DS MAX 2013 software OBJ form, with 1558 particles, the virtual hand of 3114 triangle gridding compositions and 3910 particles, 7814 triangulation network lattices Carrying out deformation simulation as a example by the Virtual Cardiac Mode become, in experimentation, model obtains and revises very convenient；Operating system For Windows 2000, based on 3DS MAX 2013, OpenGL shape library, soft at Microsoft Visual C++2012 Emulate in part development platform.

When detecting that virtual hand collides any point on virtual heart surface, given anti-clockwise torque M=10 × 10^-3Under N m effect, the regional area internal filling clockwork spring model that virtual hand is mutual with virtual heart, in interaction In, output is fed back to the reaction virtual heart real-time deformation emulation under moment of torsion effect using clockwork spring model to calculate The signal of power tactile data, as shown in Figure 2；

Under given anti-clockwise torque M effect, when virtual hand collides arbitrfary point O on virtual heart surface, virtual Lay a clockwork spring at the O of heart surface arbitrfary point, be r=1 × 10 at distance arbitrfary point O^-3Arrange a clockwork spring at m can revolve Turn mandrel points outside A₀, i.e. r is clockwork spring rotatable mandrel radius, with arbitrfary point O as initial point, ray OA₀Direction, place is X Axle positive direction, sets up XYZ space coordinate system；

Set gradually clockwork spring respectively to enclose, as shown in figures 1 and 3；Wherein, A outside the i-th-1 circle clockwork spring_i-1Point, Suspension thickness is h_i=h₁I-th circle of the clockwork spring of+(i-1) d, h₁=0.4×10^-3M, d=0.1 × 10^-3M, the i.e. i-th circle clockwork spring Spring with initial point O (0,0,0) for the center of circle, withFor the circle of radius, i=1,2,3 ... S, S are natural number；

I.e. with O (0,0,0) for the center of circle, with

R₁=r+h₁=1×10^-3+0.4×10^-3=1.4×10^-3M is the circle of radius, forms the 1st circle clockwork spring；

R₂=r+h₁+h₂=r+2h₁+ (2-1) d=1 × 10^-3+2×0.4×10^-3+1×0.1×10^-3=1.9×10^-3M is half The circle in footpath, forms the 2nd circle clockwork spring；

R₃=r+h₁+h₂+h₃=r+3h₁+3d=1×10^-3+3×0.4×10^-3+3×0.1×10^-3=2.5×10^-3M is radius Circle, formed the 3rd circle clockwork spring；

R₄=r+h₁+h₂+h₃+h₄=r+4h₁+6d=1×10^-3+4×0.4×10^-3+6×0.1×10^-3=3.2×10^-3M is half The circle in footpath, forms the 4th circle clockwork spring；

R₅=r+h₁+h₂+h₃+h₄+h₅=r+5h₁+10d=1×10^-3+5×0.4×10^-3+10×0.1×10^-3=4×10^-3m For the circle of radius, form the 5th circle clockwork spring；

The radius R of the i-th circle clockwork spring_iFormedOrdered series of numbers；

Assuming that often enclose a width of b=6 × 10 of clockwork spring^-3M, elastic modulus E=3.09 × 10⁷Pa depends on the material of flexible body Matter is the most identical；

The pilot process that calculates, last data equal round off method retain after arithmetic point 3.

Assuming that given anti-clockwise torque M=10 × 10^-3The position of N m and clockwork spring rotatable mandrel points outside A₀Place The circle at place is tangent, and under given moment of torsion M effect, if P circle clockwork spring produces torsional deflection, then before total in flexible body P circle clockwork spring is referred to as deforming check loop；

According to clockwork spring property settings: as shown in Figure 4；On front P-1 circle clockwork spring, any point is given counterclockwise Under moment of torsion M effect, the maximum twist power consumed is equal and is F₀On=0.989N, P circle clockwork spring, any point is being given Determine under anti-clockwise torque M effect, the maximum twist power consumedThe most equal, and no more than F₀=0.989N；

The decay time that front i circle clockwork spring produces torsional deflection total and needs is not more than 1ms, 1≤i≤P；And all During bar spring respectively encloses, the decay time composition that often circle generation torsional deflection needs produces torsional deflection with the 1st circle clockwork spring need to The decay time t wanted₁=10^-5S is first term, Geometric Sequence with q=1.2 as common ratio；Any point on arbitrary circle clockwork spring After the twisting resistance consumed reaches maximum twist power, its next circle starts to produce torsional deflection.

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

If on the 1st circle clockwork spring, any point is under given anti-clockwise torque M effect, the maximum twist power phase consumed Deng and all arrive F₀, then it is total to a little, on the 1st circle clockwork spring, the moment of torsion M consumed₁For:

M₁=F₀R₁=0.989×1.4×10^-3=1.385×10^-3N·m,

M₁<M,

Effective length l of the 1st circle clockwork spring₁=2πR₁,

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

On 1st circle clockwork spring, institute is a little, at the moment of torsion M of common consumption₁Under effect, torsion number of turns n produced with it₁Between Meet:

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

1st circle clockwork spring produces the decay time T that torsional deflection needs₁=t₁=10^-5s<T；

Therefore, the 1st circle clockwork spring is total to a little the moment of torsion M consumed₁< M, and the 1st circle clockwork spring generation torsion change Shape amounts to the decay time T needed₁=10^-5S < T, meets the requirement of refreshing frequency；Only when any one on the 1st circle clockwork spring The twisting resistance that point consumes reaches maximum twist power F consumed₀After, the 2nd circle clockwork spring just starts to produce torsional deflection.

If on the 2nd circle clockwork spring, any point is under given anti-clockwise torque M effect, the maximum twist power phase consumed Deng and all arrive F₀, then it is total to a little, on the 2nd circle clockwork spring, the moment of torsion M consumed₂For:

M₂=F₀R₂=0.989×1.9×10^-3=1.879×10^-3N m,

Being total to a little, on front 2 circle clockwork springs, the moment of torsion sum consumed is: M₁+M₂=1.385×10^-3+1.879×10^-3= 3.264×10^-3N·m<M=10×10^-3N m,

Effective length l of the 2nd circle clockwork spring₂=2π(R₁+R₂),

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

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

On 2nd circle clockwork spring, institute is a little, at the moment of torsion M of common consumption₂Under effect, torsion number of turns n produced with it₂Between 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 decay time T that torsional deflection needs₂=t₁+t₂=(1+q)t₁=(1+1.2)×10^-5= 2.2×10^-5S < T, T is the inverse of haptic feedback refreshing frequency here,

Therefore, front 2 circle clockwork springs are total to a little the moment of torsion M consumed₁+M₂< M, and front 2 circle clockwork springs generation torsions Deformation amounts to the decay time T needed₂=2.2×10^-5S < T, meets the requirement of refreshing frequency；Only when on the 2nd circle clockwork spring The twisting resistance of any point consumption reaches maximum twist power F consumed₀After, the 3rd circle clockwork spring just starts to produce torsional deflection.

If on the 3rd circle clockwork spring, any point is under given anti-clockwise torque M effect, the maximum twist power phase consumed Deng and all arrive F₀, then it is total to a little, on the 3rd circle clockwork spring, the moment of torsion M consumed₃For:

M₃=F₀R₃=0.989×2.5×10^-3=2.473×10^-3N m,

Being total to a little, on front 3 circle clockwork springs, the moment of torsion sum consumed is:

M₁+M₂+M₃=1.385×10^-3+1.879×10^-3+2.473×10^-3=5.737×10^-3N·m<M=10×10^{- 3}N·m

Effective length l of the 3rd circle clockwork spring₃=2π(R₁+R₂+R₃),

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

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

On 3rd circle clockwork spring, institute is a little, at the moment of torsion M of common consumption₃Under effect, torsion number of turns n produced with it₃Between Meet:

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

The decay time that front 3 circle clockwork springs generation torsional deflections need:

T₃=t₁+t₂+t₃=(1+q+q²)t₁=(1+1.2+1.2²)×10^-5=3.64×10^-5S < T,

Here T is the inverse of haptic feedback refreshing frequency,

Therefore, front 3 circle clockwork springs are total to a little the moment of torsion M consumed₁+M₂+M₃< M, and front 3 circle clockwork springs generation torsions Change shape and amount to the decay time T needed₃=3.64×10^-5S < T, meets the requirement of refreshing frequency；Only enclose clockwork spring bullet when the 3rd On spring, the twisting resistance of any point consumption reaches maximum twist power F consumed₀After, the 4th circle clockwork spring just starts to produce to reverse Deformation.

If on the 4th circle clockwork spring, any point is under given anti-clockwise torque M effect, the maximum twist power phase consumed Deng and all arrive F₀, then it is total to a little, on the 4th circle clockwork spring, the moment of torsion M consumed₄For:

M₄=F₀R₄=0.989×3.2×10^-3=3.165×10^-3N m,

Being total to a little, on front 4 circle clockwork springs, the moment of torsion sum consumed is:

M₁+M₂+M₃+M₄=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

Effective length l of the 4th circle clockwork spring₄=2π(R₁+R₂+R₃+R₄),

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

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

On 4th circle clockwork spring, institute is a little, at the moment of torsion M of common consumption₄Under effect, torsion number of turns n produced with it₄Between Meet:

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

The decay time that front 4 circle clockwork springs generation torsional deflections need:

T₄=t₁+t₂+t₃+t₄=(1+q+q²+q³)t₁=5.368×10^-5S < T,

Here T is the inverse of haptic feedback refreshing frequency,

Therefore, front 4 circle clockwork springs are total to a little the moment of torsion M consumed₁+M₂+M₃+M₄< M, and front 4 circle clockwork springs products Raw torsional deflection amounts to T between the time delay needed₄=5.368×10^-5S < T, meets the requirement of refreshing frequency；Only enclose clockwork spring when the 4th On spring, the twisting resistance of any point consumption reaches maximum twist power F consumed₀After, the 5th circle clockwork spring just starts to produce to turn round Change shape.

If on the 5th circle clockwork spring, any point is under given anti-clockwise torque M effect, the maximum twist power phase consumed Deng and all arrive F₀, then it is total to a little, on the 5th circle clockwork spring, the moment of torsion M consumed₅For:

M₅=F₀R₅=0.989×4×10^-3=3.956×10^-3N m,

Being total to a little, on front 5 circle clockwork springs, the moment of torsion sum consumed is:

M₁+M₂+M₃+M₄+M₅=(1.385+1.879+2.473+3.165+3.956) × 10^-3

=12.858×10^-3N·m>M=10×10^-3N·m

Therefore, front 5 circle clockwork springs are total to a little the moment of torsion M consumed₁+M₂+M₃+M₄+M₅> M, on front 5 circle clockwork springs Be total to a little the moment of torsion sum consumed and be not less than given anti-clockwise torque M, then the 5th circle clockwork spring is deformation check loop, is not required to To judge whether to meet the requirement of refreshing frequency again；

5th circle clockwork spring i.e. deforms the moment of torsion being total to a little consumption on check loopFor:

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

On deformation check loop namely the 5th circle clockwork spring, any point is under given anti-clockwise torque M effect, and consumption is Big twisting resistanceFor:

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

According to clockwork spring property settings: on front 4 circle clockwork springs, any point is under given anti-clockwise torque M effect, The maximum twist power consumed is equal and is all assumed to F₀Any point on=0.989N, deformation check loop namely the 5th circle clockwork spring Under given anti-clockwise torque M effect, the maximum twist power of consumptionThe most equal and be all not more than F₀=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:

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

5th circle clockwork spring i.e. deforms on check loop institute a little, at the moment of torsion M of common consumption₅Under effect, the torsion produced with it Turn-take several n₅Between 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 anti-clockwise torque M=10 × 10^-3Under N m effect, the raw torsion of common property is turn-taked several n For:

n=n₁+n₂+n₃+n₄+n₅=1.961+3.211+4.298+5.375+1.274=16.119。

Note: calculate in given anti-clockwise torque effect at the clockwork spring model using the rotatable deformation of flexible body Under, during the emulation of flexible body real-time deformation, if r, d, h₁, these parameters of b choose excessive, then the rotatable deformation of flexible body Clockwork spring model modeling method in the deformation check loop numerical value of clockwork spring just few, amount of calculation is little, and real-time is good, but becomes Shape simulated effect is the best；If r, d, h₁, these parameters of b choose too small, then the clockwork spring mould of the rotatable deformation of flexible body In the modeling method of type, the deformation check loop numerical value of clockwork spring is the biggest, and computationally intensive, real-time is the best, but deformation simulation effect Preferably；Additionally t is being set₁And t_iBetween relation time, program be should be taken into account run computer-chronograph itself hardware configuration, therefore During debugging whole program, these parameters of selection of compromising, the most repeatedly debug, so that deformation effect is more forced Very.

For verifying the implementation result of the present invention, operator touched by the handle of PHANTOM OMNI hand controller end, The deformation simulation that virtual heart is rotated by perception and control virtual hand, and the power tactile data reality that will produce in interaction Time feed back to operator.Test result indicate that: this model is simply effective, and deformation effects is true to nature, and image is smooth, and power tactile sensation is put down Surely, operator to the perception of virtual environment and mutual the most accurately and reliably, it is possible to meet the requirement of virtual teach-in interactive system.

Claims (4)

1. the clockwork spring model simulating the rotatable deformation of flexible body, it is characterised in that comprise the steps:

Step 1, sets up rectangular coordinate system in space, determines to take up an official post the clockwork spring laid at meaning point in flexible body surface, and its process is such as Under:

Step 1.1, sets up rectangular coordinate system in space,

Under given moment of torsion M effect, at the arbitrfary point O of flexible body surface, lay a clockwork spring, be at r at distance arbitrfary point O One clockwork spring rotatable mandrel points outside A is set₀, i.e. r is clockwork spring rotatable mandrel radius, with arbitrfary point O as initial point, Ray OA₀Direction, place is X-axis positive direction, sets up XYZ space coordinate system；

Step 1.2, arranges clockwork spring by circle,

Wherein, A outside the i-th-1 circle clockwork spring_i-1Point, suspension thickness is h_i=h₁The i-th circle clockwork spring of+(i-1) d, wherein h₁Being the thickness of the 1st circle clockwork spring, d is given constant and d > 0, the i.e. i-th circle clockwork spring is to be with initial point O (0,0,0) The center of circle, withFor the circle of radius, i=1,2,3 ... S, S are natural number,

The width often enclosing clockwork spring is b, cross sectional moment of inertiaElastic modulus E depends on the material of flexible body, and flexible Body material is the most identical；

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

Set position and clockwork spring rotatable mandrel points outside A of given moment of torsion M₀The circle at place, place is tangent, and in given torsion Under square M effect, if P encloses clockwork spring generation torsional deflection before total in flexible body, then P circle clockwork spring is referred to as deformation section Only circle；And P≤S, namely the number of turns of clockwork spring is at least equal to 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 consumed is equal and is F₀； On P circle clockwork spring, any point is under given moment of torsion M effect, the maximum twist power consumedThe most equal, and be not more than F₀；

Step 2.1, determines in front P-1 circle clockwork spring, the i-th circle is total to a little the moment of torsion M consumed_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 consumedFor:

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

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

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

Step 3, on calculating the i-th circle clockwork spring, institute is a little, at the moment of torsion M of common consumption_iUnder effect, 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_iBeing the effective length of the i-th circle clockwork spring, its expression formula is as follows:

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

A kind of clockwork spring model simulating the rotatable deformation of flexible body the most according to claim 1, it is characterised in that institute The decay time that stating front P circle clockwork spring generation torsional deflection total needs is not more than 1ms.

A kind of clockwork spring model simulating the rotatable deformation of flexible body the most according to claim 1, it is characterised in that institute Having in clockwork spring, often circle produces the decay time composition Geometric Sequence that torsional deflection needs, the most satisfied:

t_i=q^i-1t₁

Wherein, t_iRepresenting that the i-th circle clockwork spring produces the decay time that torsional deflection needs, q is the common ratio of Geometric Sequence, t₁For 1st circle clockwork spring produces the decay time that torsional deflection needs, 1≤i≤P.

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