CN100361829C - Wheel - Google Patents

Abstract

The present invention discloses a wheel which has the structure that m metal screw pulling and pressing springs, spiral torsion springs or torsion rod springs are arranged between a wheel rim and a wheel spoke (or a wheel hub) according to strict analysis and calculation, wherein the metal springs are connected with inflatable tyres or non-inflatable tyres in series or in parallel. The air pressure of the tyres is low, and the present invention comprises a tread bracing layer which is thickened or is made of a steel ring shell, so the wheel is flat and small. According to calculation and inference, main indexes of the wheel, such as smoothness, containment, elasticity, rolling resistance, etc., are close to or superior to the inflatable tyres and radial tyres totally. The present invention has the advantages of simple structural process and low cost.

Description

Wheel

This patent relates to wheel, especially assembles vehicle, produces elastomeric wheel with metal spring.

Common wheel is if band tyre, and elasticity is very poor; If the space that air-inflation tyre occupies is bigger.

The purpose of this patent provides a kind of wheel, and elasticity is preferably arranged when not inflating; Occupy less space during inflation.

The purpose of this patent reaches like this, is provided with the individual metal of m (＞2) (steel) spring between wheel rim that can be considered absolute rigid body and spoke (or wheel hub), and they can make wheel that elasticity is preferably arranged, and occupy less space.According to below this specification sheets with mathematics and theory of structure derive the cyclotomy trigonometric function circumferentially and ride comfort principle (being called for short the ride comfort principle) this m of design-calculated spring make wheel that ride comfort be arranged.Because metal spring does not have hysteresis, so rolling resistance is little.Because spring can have very big distortion, vehicle can not wanted pendulum spring, and spring is rectangle or round screw thread Compression and Expansion spring, has an end and supporter (wheel rim, spoke or wheel hub) captive joint at least, and the other end and supporter are fixed or be hinged.

This patent concrete structure is provided by the following drawings and example.

Fig. 1, wheel A-A section-drawing (referring to Fig. 3).

The wheel of Fig. 2, following spring tension.

Fig. 3, wheel schematic side view part.

Deformation analysis when Fig. 4, coil spring are subjected to axial force, diametral load, moment.

The A-A section of Fig. 5, Figure 10.

The rotation set of Fig. 6, coil spring end is fixed.

Fig. 7, wheel are subjected to the distortion of side force.

Fig. 8, structure of the spring bar.

Fig. 9, tire comprise some transversely arranged inflation sebific ducts.

Figure 10, wheel have four spring bars.

Figure 11, axle of spring radially press from both sides the θ angle in outer end and wheel.

Figure 12, the common airtyred wheel of use.

Figure 13, the deformation analysis of spring bar 19 in wheel.

One of setting of Figure 14, torsion bar spring.

Two of the setting of Figure 15, torsion bar spring.

The distortion in wheel of Figure 16, torsion bar spring.

Be not full of sizing material between Figure 17, the transversely arranged sebific duct.

Figure 18, bracing ply 23 ' are hard-shell type.

Figure 19, carcass 23 are made arch form.

Figure 20, ground connection center carcass horizontal wall and wheel rim 4 are tangent, and gap 28 disappears.

Fig. 1,3 wheels comprise the individual metal that radially is provided with along wheel of m (=4) (steel) cylinder (or rectangular column) tension and compression coil spring 3, spring silk section circle (also can rectangle).Spring is along circumferentially uniform, i.e. branch wheel circumference such as m.The bearing coil diameter is less, and the outer end circle is captiveed joint with wheel rim 4 with end circle annex (screw 9, nut 10, packing ring 13 (the best surfaces cutting increases matching surface)).Inner circle is used with quadrat method and is connected with spoke 2, and spoke 2 usefulness screws 12 are connected with wheel hub 11.The inner wheel hub 11 that also can directly be connected that encloses.Wheel rim 4 can but comprise the back-up ring of fixing with screw 87.Also can make wheel rim 4 internal diameters less than spoke 2 external diameters by Fig. 2.This structure is bigger by the calculating wheel lateral rigidity of back.With wheel rim 4, the structure that spring 3, spoke 2 are formed is universal spring (seeing the back for details), and the space that it occupies is less.Figure 11 spring (axis) 3 also can radially press from both sides the θ angle at the wheel of outer end and free state.Constant right when the bearing coil diameter, packing ring 13 can change rotation set 14 among Fig. 6 into, and rotation set 14 also can be divided into two parts 16,17 (Fig. 8).The 16th, spiral cover; The 17th, cock.Spiral cover and cock can need only one of them.End circle and supporting mass (wheel rim etc.) are hinged.With Fig. 6 is example, unloads screw 9, and nut 10 separates spoke 2 and rotation set 14 springs 3.Then screw 9 nuts 10 reverse (screw head is placed in the rotation set 14) are tightened on the rotation set, and make screw 9 extend one section growth bar 21 (see figure 8), when bar 21 usefulness pins 15 are hinged with supporting mass, hold circle promptly to be hinged with this supporting mass.The supporting mass that can be hinged is separately enclosed at two ends, and when the spring ratio of height to diameter was big, the head of two screws 9 is extensible to be guide cylinder and pilot bar 18.Spring is called spring bar 19 (Fig. 8) with the end circle annex with two extension rods 21.Two pin width between centerss 1 are the length of spring bar.Figure 10 is the wheel that four spring bars 19 are hinged between wheel rim 4 spokes 2.The spring 3 of Fig. 1 also can change an end into hinge.The spring bar 19 of the m on the wheel two ends hinge also can change a m torsion spring spare (Figure 14,15) into.Torsion spring spare is the arm 24 long lsin θ of being ₆Torsion-coil spring (or torsion bar spring) be lcos θ with long ₆Connecting rod 26 form connecting rod 26 1 end hinge wheel rims 4, other end hinge torsion spring body 25 (Figure 14) or arm 24 (Figure 15), the torsion spring arm 24 among Figure 14 spoke 2 that is hinged again.Torsion spring body 25 hinge spokes 2 among Figure 15 (are that the relative spoke 2 of torsion bar axial location is fixed.The torsion bar two ends all are hinged on spoke 2).Connecting rod 26 radially presss from both sides θ angle (Figure 16) in outer end and wheel.Can there be auxiliary mechanism that torsion bar is not subjected in case of necessity or be subjected to moment of flexure less, only be subjected to moment of torsion.The torsion bar section can be circular rectangle or lamination, lamination shock damping action arranged.Can between wheel rim and spoke, damping be set for whole described universal elastic elements.Below universal spring (also being a kind of wheel) is done computational analysis.

Fixing upper end, lower end is the cantilever coil spring freely, and the upper end is subjected to moment M, diametral load F _r, the deformation analysis of axial force F (Fig. 4, axial force are pulling force, if pressure F is a negative value, spring is with representing when gauge rod) is as follows.Wherein D is a spring mean coil diameter, and d is the spring filament diameter.N is a number of active coils.H is an operating altitude, H _oBe free height, E is a modulus of elasticity, and G is a shearing elastic modulus, and μ is a Poisson's ratio, B=HEA/[4 (2+ μ)] be the equivalent bending stiffness, S=HEA/D ²Be the equivalent shear rigidity, F '=Gd ⁴/ (8D ³N) be axial push-pull rigidity.A=d wherein ⁴/ (8Dn).Fig. 4 regards spring as and works as gauge rod, and the crooked corner λ that produces is arranged after the cross-sectional plane before the distortion loads, and shears the corner φ of the relative bar line of centers in cross section that produces, establish the cross section far from the lower end apart from x, be subjected to a power F, shearing force-F λ+F _r, moment of flexure-Fy+M+F _r(H-x), φ=(F λ+F _r)/S, d λ/dx=[-Fy+M+F _r(H-x)]/B, dy/dx=-(φ+λ)=-(1-F/S) λ-F _r/ S.By the above various differential equation d that has ²Y/dx ²-(F/B) (1-F/S) y=-(1-F/S) [M+F _rR (H-x)]/B.Make q ²=(F/B) (1-F/S).F＞0 an o'clock q is an arithmetic number; F＜0 o'clock is imaginary number q=|q|i.Separating of this equation is y=C ₁Sinh qx+C ₂Coshqx+[M+F _r(H-x)]/F.By boundary condition x=0, λ=0; X=H, y=0 has C ₁=(F _r/ Fq) (1-F/S); C ₂=(F _r/ Fq) (1-F/S) tanhqH-(M/F) SechqH.Upper end radial deformation f _r=y| _X=H=(F _rH/F) [1-tanhqH/ (qH)]+F _rH/S+ (M/F) is (1-SechqH): angular deformation γ=λ | _X=H=(1-F/S) ^-1[-(F _r/ F) (1-F/S) (1-sechqH)-(Mq/F) tanhqH] ... (2).Make η ₁=3 (qH-tanhqH)/(qH) ³, η ₂=2 (1-SechqH)/(qH) ², η ₃=tanhqtH (qH) ... (1).ξ ₁＝η ₁(1-F/S)，ξ ₂＝n ₂(1-F/S)。η (general reference η ₁, η ₂Deng) be axle power influence coefficient, all be the real variable function, because of order | qH|=v, v=qH when F is pulling force, η ₁=3 (v-tanhv)/v ³, η ₂=2 (1-sechv)/v ², η ₃=tanhv/v.When F is pressure, iv=qH, η ₁=3 (tanv-v)/v ³, η ₂=2 (1-secv)/v ², η ₃=tanv/v.So above-mentioned distortion (2) formula is f _r=F _rH ³ξ ₁/ (3B)+F _rH η ₃/ S+MH ²ξ ₂/ (2B), γ=-FrH ²ξ ₂/ (2B)-MH η ₃/ B.Make u=q ²H ², as u=-π ²/ 4 o'clock η=infinities are unstability radially, and when u=0, η equals 1.If the approximate η that gets ₁=η ₂=η ₃=η, 1-F/S=1, the above-mentioned f that is deformed into _r=[F _rH ³(3B) ^-1(1+3BH ^-2S ^-1)+MH ²(2B) ^-1] η, γ=-[F _rH ²(2B) ^-1+ MH/B] η.As seen η is the influence of a power F to distortion.With (2) formula to F _r, M has solved.F _r＝2B[Δ(1+μ)HD ²] ^-1(2+μ)(f _rη ₃-0.5γHξ ₂)，M＝2(2+μ)B[Δ(1+μ)HD ²] ^-1{[H ²ξ ₁/3+0.25(2+μ) ^-1D ²η ₃]γ-0.5f _rHξ ₂}。Δ=2 (2+ μ) (1+ μ) wherein ^-1(H/D) ²(ξ ₁η ₃/ 3-0.25 ξ ₂ ²)+0.5 (1+ μ) ^-1η ₃ ². substitution B=[32Dn (2+ μ)] ^-1HEd ⁴=0.5 (2+ μ) ^-1HD ²F ' (1+ μ), S=(8D ³N) ^-1HEd ⁴=2 (1+ μ) HF ', E=2 (1+ μ) G.And make f ₁=f/H ₀, a ₁=1-0.5 (1+ μ) ^-1, a ₂=0.5 (1+ μ) ^-1, a ₃=2 (1+ μ) ^-1(2+ μ).Can be exchanged for the flat f of axial friendship shape to axle power F ₁As (qH) ²=a ₃(H _o/ D) ²f ₁(1+a ₁f ₁), H (1-F/S)=H _o(1+a ₁f ₁), Δ=a ₃(H _o/ D) ²[3 ^-1(1+f ₁) (1+a ₁f ₁) η ₁η ₃-0.25 (1+a ₁f ₁) ²ξ ₂ ²]+a ₂η ₃ ², F=F ' H _of ₁, F _r=F ' [g ₃f _r-0.5H _o(1+a ₁f ₁) g ₂γ], M=F ' { [3 ^-1(1+f ₁) (1+a ₁f ₁) g ₁H _o ²+ a ₃ ^-1a ₂D ²g ₃] γ-0.5H _o(1+a ₁f ₁) g ₂f _r.G wherein ₃=Δ ^-1η ₃, g ₁=Δ ^-1η ₁, g ₂=Δ ^-1η ₂... (3) now come the wheel of scaling system 3 with these, the ground connection position is subjected to normal force Q, horizontal force Q φ. φ=tan θ ' is a traction coefficient. under the load, the wheel rim 4 that is considered as rigid body is along β angle miles of relative movement h, the center arrives O ' point, and rotates the γ angle. and upright fixed coordinate system ZO ' Y, the Z axle overlaps OO '. first axle of spring and Z axle clamp α angle before loading, make r=2 π/m, i spring and Z axle clamp angle α _iThe r.i=1 of=α+(i-1), 2...m.R is the distance of spring outer end to wheel rim 4 centers. then there is axial deformation f spring i outer end _i=hcos α _i, radial deformation f _Ri=R γ+hsin α _i, angular deformation γ _i=γ .... (4) these distortion produce axial force F in the spring outer end by (3) formula _i, diametral load F _Ri, moment M _i, get them at the Z axle, the projection on the Y-axis and to the square of O '. again to i=1 to the m summation, be called circumferentially and, m represents with ∑, should equal external force Q, φ Q respective projection and moment. be equation of equilibrium ∑ m (F _iCos α _i+ F _RiSin α _i)=Qcos β-φ Qsin β, ∑ m (F _iSin α _i+ F _RiCos α _i)=Qsin β+φ Qcos β, ∑ m (F _RiR+M _i)=φ QR. transform (4) makes and f ₁Relevant amount g (general reference g _iDeng), it is relevant with angle α that Δs etc. become, and is the function of α. at α=α _iThe time its functional value be designated as g _iΔ _iDeng. arrangement back balance equation is .F ' b ₁+ F ' γ b ₂=(1+ φ ²) ^0.5Qsin (β+θ '), F ' b ₃+ F ' γ b ₄=(1+ φ ²) ^0.5Qcos (β+θ '), F ' b ₅+ F ' γ b ₆=φ QR. is h wherein ₁=h/H ₀, b ₁=∑ m[0.5H ₀h ₁Sin2 α _i(1+g _3i)], b ₂=∑ m{cos α _i[Rg _3i-0.5H ₀(1+a ₁h ₁Cos α _i) g _2i], b ₃=∑ m[H ₀h ₁(cos ²α _i+ g _3iSin ²α _i)], b ₄=∑ m{sin α _i[Rg _3i-0.5H ₀(1+a ₁h ₁Cos α _i) g _2i], b ₅=∑ m{H ₀h ₁Sin α _i[g _3iR-0.5H ₀(1+a ₁h ₁Cos α _i) g _2i], b ₆=∑ m{-H ₀(1+a ₁h ₁Cos α _i) g _2iR+3 ^-1H ² _o(1+h ₁Cos α _i) (1+a ₁h ₁Cos α _i) g _1i+ (R ²+ a ₂D ²/ a ₃) g _3i... (5a) this equation can be to beta, gamma, and F ' solves, β=arcsin[d ₁/ (d ² ₂+ d ² ₃) ^0.5]-arcsin[d ₃/ (d ² ₂+ d ² ₃) ^0.5], F '=φ QRd ₄/ d ₁, γ=(d ₁-b ₅d ₄)/(b ₆d ₄). d wherein ₁=φ R (b ₂b ₃-b ₁b ₄), d ₂=[b ₃b ₆-b ₄b ₅+ φ (b ₁b ₆-b ₂b ₅)], d ₃=φ (b ₃b ₆-b ₄b ₅)-(b ₁b ₆-b ₂b ₅), d ₄=b ₂(cos β-φ sin β)-b ₄(sin β+φ cos β) ... (5). the beta, gamma that solves, F ' they all are the functions of α. the interval of α is [0,2 π/m]. appoints and gets such function,, make γ as γ ₀=γ | _α=0, the peak excursion of γ in the interval is designated as max| γ-γ ₀|, then maximum relativity shift δ=γ ₀ ^-1Max| γ-γ ₀|. when δ less than given δ ₀The time wheel ride comfort is arranged. can give H in advance during design ₀, D, f ₁, parameters such as m are calculated beta, gamma by (5) formula, and F ' is not if their relativity shift is all less than δ ₀Give number (counting m) up to all less than δ in advance with regard to all or part of modification as increasing spring ₀.

Be approximation method below, get function η ₁=η ₂=η ₃So=η is g ₁=g ₂=g ₃=g=η ₃ ^-1[3 ^-1a ₃(1+f ₁) (1+a ₁f ₁) y-0.25a ₃(1+a ₁f ₁) ²Y+a ₂] ^-1. y=(H wherein ₀/ D) ²... (8a). make e=a again ₃Y/12+a ₂, c=a ₃Y (2-a ₁So)/(6e). g=[e (1+cf ₁) η ₃] ^-1... (8b) by (1) formula, wherein η ₃ ^-1=(qH) ctgh (qH)=1+u/3+u ²/ 45+2u ³/ 945+.... exists | u|＜π ²Convergence in/4 o'clock can only be got 3. (1+cf ₁) ^-1=1-(cf ₁)+(cf ₁) ²... the condition of convergence is | f ₁|＜c ^-1. make w=a ₃Y is with u=wf ₁(1+f ₁) and two progression substitutions (8b) formula, and make a ₄=[c ²-(wc)/3+ (wa ₁)/3+w ²/ 45]. g=e is arranged ^-1[1-(c-w/3) f ₁+ a ₄f ₁ ²+ ...+(1) ⁿc ^N-2a ₄f ₁ ⁿ+ ...] .... (8). the item number that progression (8) is got is to be determined by data of giving in advance and error requirements. for the sake of simplicity, be example to get 2. be g=e ^-1+ a ₅f ₁, a wherein ₅=e ^-1(w/3-c) .g is f ₁Multinomial. substitution formula (4) and (5a) back ∑ m summation be exactly that t trigonometric function multinomial sued for peace. utilize conversion sin ²α=1-cos ²α knows that its general term is cos ⁿα _i, sin α _iCos ^N-1α _i, point out ∑ m (sin α when m＞t below n＜t. _iCos ⁿα _i∑ m (cos when)=0.n is odd number ⁿα _i)=0; ∑ m (cos when n is even number ⁿα _i)=constant. as ∑ m (cos ²α _i)=m/2. is irrelevant with α. thereby the b (b that calculates by (5a) _iDeng) skew is zero. because cos ⁿα _iDeng can turning to a times angle function cosk α through long-pendingization and difference _i, sink α _iThe first power of (k is less than n+1). and ∑ m (sinn α _i), ∑ m (cosn α _i), n＜m is except ∑ m (cos0 α _iAll the other are zero for)=m. reason is ∑ m (sin α when n=1 _i), ∑ m (cos α _i) be the projection of m radius vectors sum on coordinate axle in length and breadth of branch unit circle such as m, so be zero. when n＞1, establish m, n has greatest common divisor p (can be 1), m=pq, ∑ m (cosn α _i)=p ∑ q (cos α _j) and formula ∑ q (cos α _j) be q vector of q Equal round in the former m vector. so ∑ m (cosn α _i)=0. is ∑ m (sinn α in like manner _i)=0. is sinn α _i, cosn α _i, i=1...m, n＜m are the cyclotomy trigonometric function, this character be called the cyclotomy trigonometric function circumferentially and ride comfort. be called for short the ride comfort principle. the b in m＞3 o'clock (5a) in view of the above ₁=b ₄=b ₅=0.b ₂=mh ₂[0.5Ra ₅-0.25H _o(a ₅+ a ₁e ^-1)], b ₃=0.5mH ₀h ₁(1+e ^-1), b ₆=me ^-1[H ₀R+3 ^-1H ₀ ²+ R ²+ a ₂D ²/ a ₃]. solve β=-θ '. represent at O ' at Q .F ' on the resultant direction of φ Q=2 (1+ φ ²) ^0.5Q/[mH ₀h ₁(1+e ^-1)] .... (10), γ=φ he (1+e ^-1)/[2Rd ₅] .... (11) are d wherein ₅=1-H ₀/ R+3 ^-1(H ₀/ R) ²+ (D/R) ²a ₂/ a ₃. substitution F '=Gd ⁴/ (8D ³N), the radial rigidity of wheel is mGd ⁴(1+e ^-1)/(16D ³N), tangential rigidity is φ Q/ (R γ)=mGd ⁴d ₅e ^-1/ (8D ³N), lateral rigidity is mGd ⁴(e ^-1-0.5a ₅h ₁)/(8D ³N). (the wheel lateral rigidity of Fig. 2 is mGd ⁴(e ^-1+ 0.5a ₅h ₁)/(8D ³N)). as seen each rigidity all with e ^-1Relevant, and e ^-1Being that ratio of height to diameter y with spring reduces and increases, can not too big .Q be fully loaded load so be used as the spring ratio of height to diameter of wheel, makes Q ₀The safety factor of=Q/m. spring should be considered braking, reverse stress etc.Because reverse stress has big safety factor, braking can no longer be considered. maximum shear stress is in inboard, Fig. 1 spring lower end, and the moment of torsion that produces this stress is estimated as Q ₀d ₆. d wherein ₆=(1+e ^-1) ^-1[0.5d ₅ ^-1(1+e ^-1) φ H ₀+ (D ²+ (H ₀/ e) ²) ^0.5]. strength condition is τ _Max=16Kd ₆Q ₀/ (π d ³)＜[τ] .... pre-data D, d, the H of giving during (12) specific design _o, h ₁, n, R, wherein H ₀＞nd+h calculates y, w, c, e, a ₄, a ₅, d ₅, d ₆, etc., and check | u|＜π ²/ 4. determine the item number t that expansion equation (8) formula of g should be got, the formula Q that derives by (10) formula of substitution then ₀=Q/m=Gd ⁴H (1+e ^-1)/(16D ³N). substitution as a result (12) formula is carried out strength check, comprise that then (m＞t+1) wheel of individual this spring becomes a series of wheel to m, and their rated load is mQ if satisfy (12) formula _o. deflection all is h, and when m was big, m spring also can be arranged in 2 or a plurality of circumferential circle. different springs and different deflection h can have different wheel subfamilies, and two wheels in the different series can have identical load.

Example: H ₀=D=70mm, d=9mm, n=4, h=23mm, R=350mm, φ=0.3, [τ]=0.4 σ _b=680MP _a. design process, 1, | u|＜π ²/ 4.e=0.679.d ₅=0.81.d ₆=62.3mm.D/d=7.78.K=1.13.Q ₀=1360N. τ _Maxa=669MP _a＜[τ]. when m=4, Q=5440N, during m=8, Q=10880N.

When the spring outer end when radially pressing from both sides θ angle (Figure 11), corresponding (4) formula should be f _i=R γ sin θ+hcos α _i, f _Ri=R γ cos θ+hsin α _i, γ _i=γ can get design formula with identical method then. and when spring outer end and wheel rim 4 was hinged (referring to Fig. 5), as long as make moment M=0 among Fig. 4, identical derivation can get design formula. as spring axial stiffness F ' and f ₁When relevant, F ' is f ₁Power series, getting behind the finite term also can be with ride comfort principle derivation formula. when both ends of the spring all is hinged on supporting mass and outer end and radially press from both sides the θ angle, the distortion of spring and the derivation of equation are Figure 13, setting up of the stressed and coordinate axle ZO ' Y of wheel rim 4 is identical with Fig. 3. and spring bar (length l) 19 outer ends are along the tangential direction and Y-axis (level) the folder α angle of wheel, (spring bar 19 and α should have subscript i here, succinct in order to narrate, wouldn't add i, when ∑ m sues for peace, add again). loading the outer end, back has tangential displacement t=R γ, Z becomes long l to displacement h ₁Dotted line 19 ', can think that first deformability tangential is length u, Z is to being deformed into l again ₁, θ ₁Be u and tangential, θ ₂For u and Z to angle. θ ₃Be l ₁With Z to angle. by figure, u ²=l ²+ t ²-2ltsin θ, ucos θ ₁=t-lsin θ, sin θ ₂=-cos (θ ₁+ α), l ₁ ²=u ²+ h ²+ 2uhsin (θ ₁+ α), l ₁Sin θ ₃=usin θ ₂, l ₁Cos θ ₃=h-ucos θ ₂, cos (θ ₃+ 0.5 π-α)=-sin (θ ₃-α). with reference to the fwd mark equation of equilibrium is arranged.∑ m[F ' (l _1i-1) sin θ _3i=-F ' ∑ m[(l _1i-1) l _1i ^-1Ucos (θ ₁+ α _i)]=(1+ φ ²) ^0.5Qsin (β+θ '). ∑ m[F ' (l _1i-1) cos θ _3i]=F ' ∑ m{ (l _1i-1) l _1i ^-1[h+usin (θ ₁+ α _i)]]=(1+ φ ²) ^0.5Qcos (β+θ '). ∑ m[F ' (l _1i-1) Rcos (θ _3i+ 0.5 π-α _i)]=∑ m{F ' (l _1i-1) l _1i ^-1[hsin (θ ₁+ α _i) cos θ ₁-hcos (θ ₁+ α _i) sin θ ₁+ ucos θ ₁]=φ QR. (l wherein _1i-l) l _1i ^-1=l-ll _1i ^-1=l-(u ²+ h ²)- ^0.5[1-0.5a+0.375a ²... .], a=2uh (u ²+ h ²) ^-1Sin (θ ₁+ α _i) ... .. (8 '). when progression in the bracket is got the t item and is satisfied error requirements. according to the wheel of smooth-going principle design should m＞t. because 1 than h, so big many of t are so desirable 2. m＞2. solve at this moment β=-θ ' .Q _o=Gd ⁴H (8D ³N) ^-1[2 (1+ φ ²) ^0.5+ φ h/ (lsin θ)] ^-1.... (10 ') .t=8 φ Q ₀D ³N/ (Gd ⁴Sin ²θ) .... (11 ') have similar design process .1, give data D, H in advance ₀, n.h, Q, l. obtains Q by (10 ') ₀, carry out strength check.τ _Max=8KDQ _o(π d ³) ^-1＜[τ]. if satisfy strength condition.Serial wheel is promptly arranged, m=3,4 ... know θ=0 o'clock by (11 ') formula, the t=infinity is so should be not equal to 0 by θ.Can ask the θ angle according to the tangential rigidity of the wheel of giving in advance conversely.In (10 ') θ change into-θ after, Q _oDifference illustrates that wheel advances, radial rigidity difference when retreating, as require the forward-reverse radial rigidity identical, and m=2k spring bar 19 can be set, wherein get the θ angle for k, get-the θ angle for k in addition, at this moment Q _o=Gd ⁴H/[16D ³N (1+ φ ²) ^0.5]. (when θ=π/4 (Figure 10), if get (1+ φ ²) ^0.5=1, radial cut is arranged to equal stiffness, i.e. Q/h=φ Q/t=Gd ⁴/ (16D ³N), Zui Xiao m=4).Figure 16 is distortion of torsion bar spring wheel and derivation graph.The axis of torsion bar 25 is fixed on the spoke 2, i.e. the two ends of the torsion bar spoke 2 that all is hinged.Long is lsin θ ₆ Torsion bar arm 24 through long lcos θ ₆Connecting rod 26 link to each other with wheel rim 4.Arm 24 vertical links 26 during free state.Connecting rod 26 and Z axle clamp α angle radially press from both sides the θ angle in outer end and wheel.Loading rear linking rod 26 outer ends has along wheel tangential displacement t, and Z makes arm 24 rotational angle θ to displacement h ₄=(tsin θ ₆+ hcos α)/(lsin θ ₆). connecting rod 26 rotational angle θ ₅=(tcos θ ₆-hsin α)/(lcos θ ₆). angle θ ₄Make torsion bar produce torque GJ _pθ ₄/ L, wherein J _p, L is the torsion bar utmost point moments of inertia and effectively long, makes a=θ ₄-θ ₅=(h/l) (sin α/cos θ ₆+ cos α/sin θ ₆), connecting rod 26 stressed F=GJ _pθ ₄/ (L1sin θ ₆Cosa), make b=θ ₅Following three useful sin α are arranged, the power series of cos α.Cos ^-1A=1+0.5a ²+ (5/24) a ⁴+ ... .sin θ ₅=b-b ³/ 6+b ⁵/ 120....cos θ ₅=1-0.5b ²+ b ⁴/ 24.... (8 "). because connecting rod 26 becomes α-θ with Z axle and angle radially ₅, θ-θ ₅So,, make a ₇=GJ _p/ (Llsin θ ₆) the back balance equation is ∑ m[F _iSin (α _i-θ _5i)]=a ₇∑ m{ θ _4iCos ^-1a _iSin (α _i-θ _5i)=(1+ φ ²) ^0.5Qsin (β+θ '). ∑ m[F _iCos (α _i-θ _5i)]=a ₇∑ m{ θ _4iCos ^-1a _iCos (α _i-θ _5i)=(1+ φ ²) ^0.5Qcos (β+θ '), ∑ m[F _iSin (θ-θ _5i) R]=a ₇∑ m{ θ _4iCos ^-1a _iSin (θ-θ _5i) the following factor in the R}=φ QR. equation of equilibrium is deployable, sin (α-θ ₅)=sin α cos θ ₅-cos α sin θ ₅, cos (α-θ ₅)=cos α cos θ ₅+ sin α sin θ ₅, sin (θ-θ ₅)=sin θ cos θ ₅-cos θ sin θ ₅Respectively get this three formula of finite term substitution according to three progression of error requirements in will (8 "), the smooth-going principle that again this three formulas substitution equation of equilibrium just can be used the cyclotomy trigonometric function determines the value of m.As an example for getting cosa=1 for simplicity, cos θ ₅=1, sin θ ₅So=0. solve β=-θ ', h=2 (1+ φ ²) ^0.5QLl ²Sin θ ₆/ (mGJ _p) .... (10 "), t=φ QLl ²Sin ²θ ₆/ (mGJ _pSin θ) .... (11 ") design process is the same.Here m＞2.Figure 11, the θ angle in 13,16 can be used for regulating wheel radially with tangential rigidity ratio, promptly when radial rigidity is very little, keep tangential rigidity not too little.

Below all be to disregard the formula that wheel rim 4 distortion are derived, when taking into account wheel rim 4 distortion, also can provide the method for designing of spring, be general ride comfort principle with similar principle (or test).

Universal spring links to each other with tire 5 and can produce necessary impression and ground pressure distribution.During series connection (Fig. 1,9,17,19), tire and is combined on the wheel rim 4; (Figure 12,18) is arranged in the common air-inflation tyre 5 when in parallel, above and between carcass gapped 28.Can reduce inflation pressure, because universal spring has the added resilience effect, and the Runflat effect.Common airtyred back-up ring, retainer is fixed on spoke 2 and tyre bead on the main wheel rim 22 together.Fig. 1 air-inflation tyre is the flat annular arrangement shell of making of soft strengthening course 23 of section, and the upper wall of shell is subjected to circumferential pulling force bigger, additional bracing ply 23 '.In fact Fig. 1 tire is the very big radial-ply tyre of a kind of constraint Fr.It is for fear of weak tyre bead occurring that carcass is made the annular arrangement shell.Bracing ply 23 ' and carcass 23 adhesion one among Fig. 1 for fear of weak tire shoulder occurring, also can be made split, and promptly the tyre surface 5 with bracing ply 23 ' adhesion is detachable removable treads.Detachable removable tread also can be regarded cover tire as, and annular arrangement shell 23 is regarded the inner tube of a tyre of being with reinforcement as.These argumentations can be used for improving existing radial-ply tyre.Promptly 1, between wheel rim and spoke (or wheel hub), spring 3 is set, increase the traveling comfort that uneven road surface travels.2, carcass is made annular arrangement shell (carcass is an osed top on the cross sectional drawing).3, the carcass of annular arrangement shell and belt layer tread split, as the inner tube of a tyre of band strengthening course.Can simplify technology; And advantage with tubeless tires.Also can change the carcass below the bracing ply 23 into several little carcass arranged side by side, each little carcass is the constraint F that a sebific duct ring 23 (Figure 17) can reduce radial-ply tyre _rSeveral sebific duct rings 23 also can be that long length hose helix on wheel rim is entwined.Also can make circular (Fig. 9) substantially to sebific duct, and be with or without cushion layer between tyre surface.The sebific duct strengthening course is stressed little, but cost-cutting also has Runflat.The size air chamber all can change water-filling wet goods liquid.Can be full of sizing material or not be full of sizing material as Fig. 9 between adjacent sebific duct as Figure 17.Consider that Fig. 1,9,17, sidewall deflection are bigger, can make arch form (referring to Figure 19) to carcass 23, ground connection place amount of deflection is not produced by the sidewall entirely.(Figure 12) should reduce inflation pressure for the universal action of the spring of performance during parallel connection, can increase the bending stiffness I of bracing ply 23 ' in circumferential plane for the ground pressure distribution was constant after air pressure was reduced _y(increase the bracing ply gross thickness, cry thicken) as between the wirecord fabric bracing ply, filling out the hard sizing material.Air pressure reduces to zero during extreme case, becomes non-inflatable tyre.Bracing ply 23 ' this moment (Figure 18) can be endless belt or the toroidal shell (abbreviation hard-shell type) that solid-state high modulus material (metal, plastics, steel commonly used) is made.Radial position is with carcass cord steel wire or similarly structural constraint (in detail can referring to the firm lining 4 among parts among the patent US4111249 34 or the CN2251501).Outer in case of necessity sticking hard glue-line 27 makes the rigidity continuous transition.Sticking or lining is example (also can outside) carcass 23 in.Because of the constraint of carcass 3 and wheel rim 4, produce flexural deformation (Figure 20 solid line bracing ply 23 ') applying ground 29 as laminated spring under the Radial load, core falls in the groove of wheel rim 4 (Figure 18 dotted line), makes gap 28 disappear universal deflection of spring.The straight portion of Figure 20 bracing ply 23 ' makes the ground pressure face strengthen as backing plate, does not need very big I _yStraight portion bends to dotted line 23 ' when running into little obstacle, and wheel rim 4 is pushed up dotted line position, (Figure 20), and envelope (containing) property is arranged.Because the ring spring 23 ' that is deformed to Figure 20 state has diminished and air pressure is zero in grounded neutral place radial rigidity, the described pardon of deducibility is poor unlike common air-inflation tyre (oblique or radial).Reduce more than 60% than common air-inflation tyre owing to zero air pressure carcass 23 cords are stressed in addition, so carcass can be very thin, thereby aspect ratio can reduce, and also is applicable to cascaded structure (Fig. 1, this moment, gap 28 can not disappear).Structure when Fig. 1 is connected also has similar description, and under the unmodified Radial load, bigger air pressure can be kept on the carcass lower wall separately makes load complete all by series connected air bellow carcass.Be equivalent to common air-inflation tyre, so aspect ratio can not be too little, the effect of universal spring is the elasticity that increases conventional tyre.Along with air pressure reduces, but non-vanishing, by Figure 20, ground connection center carcass horizontal wall and wheel rim 4 are tangent, and gap 28 disappears.Thereby inflate carcass and no longer play elastic reaction, but still have air pressure to produce necessary ground pressure in the front and rear part, point of contact.Because the low aspect ratio of air pressure can be little, carcass can approach and make pardon good, thereby tire can normal operation.Also available rigid plastic of non-inflatable tyre or steel thin-walled ring (hard-shell type) do carcass 23 be combined on wheel rim 4 (Figure 19) or the main wheel rim 22.23 liang of flanges of toroidal shell (also being tyre bead) close fit produces tangential rigidity on wheel rim, load lower house flexural deformation in section (dotted line).The non-inflatable tyre of connecting in addition can also be a band tyre, comprises common band tyre; Or made middle cavity with big (as the poly-urethane rubber) rubber of pulling strengrth, or do not have cavity but the foam sponge shape to increase the band tyre of impression area.Perhaps have roughly as the shape of Figure 12 (air-inflation tyre) but do not have the reinforcement cord, tyre sidewall is thicker and harder, be rubber-type carcass or rubber-type bracing ply, the tire that adds up to except that Fig. 1,9,17 with universal spring serial or parallel connection has common inflation (radial, oblique band bundle oblique) or non-inflatable tyre, and inflation or non-inflatable tyre with air-inflation tyre shape, its bracing ply 23 ', carcass 23 one of or two are rubber-types, cord type or hard-shell type.

Be what time replenishing below to preamble.1; K in the 4th page of (12) formula is the curvature coefficient of correction.2; In Fig. 6 and Fig. 8, when not waiting, the pitch of pitch on spiral cover 16 cocks 17 rotation sets 14 and spring 3 can not want nut 10 yet.3; Screw 9 among Fig. 1 and Fig. 6 Fig. 8 also can with spring 3 one, promptly be that spring 3 spring silks in beach extend and process screw thread in the above, at this moment can not want spiral cover 16 cocks, 17 rotation sets 14 and packing ring 13, as long as nut 10 just can be fixed on spring 3 on the wheel rim 4.4; During various universal spring rolling work, drawing-pressing spring

3 are subjected to symmetrical cycle alternation axial force, and torsion spring spare is subjected to the symmetrical cycle alternate torque.Can be with changing spring 3 free heights (therefore Gao Jing

Than changing) etc. method make each spring 3 that identical initial tension or initial pressure (existing axial tension under the stand under load free state neither all be arranged

Or pressure).Make each torsion spring spare that identical first moment of torsion all be arranged with methods such as changing torsion spring spare size 1.Original like this symmetrical cycle

The alternation axial force just becomes asymmetric or pulsation repeat power.Distortion by these internal force generations just is designated as fBelow be example with Fig. 3

Explanation.Because initial set is arranged f, the axial deformation f in corresponding spring outer end distortion (4) formula _iShould change into: f _i= f+ hcos α;

。Can get design formula with identical derivation method then.Obviously the gained result will with initial set fRelevant.When fBigger, spring

Axially only be subjected to the contact stress at pulling force (or pressure) and whole end circle and supporting mass dividing range place to have only tensile stress (or compression effort all the time

) time, the fixing or hinge of end circle and supporting mass connects and can adopt known stretching (or compression) side that bearing coil is fixed or hinge connects

Method connects.This connection of compression spring can be referring to mechanical press " mechanical engineering manual " second edition Course Exercise in Machinery Elements Design volume the

The 10-23 page or leaf, Figure 10 .2-5, or referring to this Figure of description 12 (intermediate portion also is the A-A section-drawing of Fig. 3), end circle annex

Has only a flange that has just connected with supporting mass (regarding no spiral fluted cock as).Structure is very simple.5; (4) formula, (5) formula and (

10) R in the formula is wheel rim 4 inside radiuss among Fig. 3, promptly during free state spring 3 outer ends to the distance at wheel rim 4 centers, to Fig. 3 structure

The derivation of wheel ignored the tire height, thereby horizontal force φ Q is φ QR to the square of O ' in the equation of equilibrium.When considering the tire height

The time, horizontal force φ Q should change φ QR ' into to the square of O ', and R ' is that ground-surface height is arrived at the wheel rim center, and R ' is greater than R, and R ' is less than outside the tire

Radius is because tire has distortion under the diametral load effect.Because moment of face (φ QR ') increases, wheel rim 4 corner γ also increase, and adopt

Less spring ratio of height to diameter perhaps adopts the structure of θ among Figure 11 ≠ 0 can increase the tangential rigidity of wheel, reduces corner γ.

Claims (8)

1. a wheel comprises wheel hub, wheel rim (4), it is characterized in that between wheel hub and wheel rim (4) according to the cyclotomy trigonometric function circumferentially and the ride comfort principle be provided with m metal spring (3), each spring has at least an end to be set to captive joint.

2. wheel according to claim 1, it is characterized in that described spring is rectangle or cylindrical spiral Compression and Expansion spring, spring silk section circle or rectangle, the other end and spoke or wheel hub are set to fixing or are hinged, axle of spring radially presss from both sides the θ angle in outer end and wheel during free state, θ=0 or θ ≠ 0.

3. wheel according to claim 2, it is characterized in that described Compression and Expansion spring is extension spring or compression spring, be spring shaft to because there is initial stress only to be subjected to pulling force or pressure, hold fixing of circle and supporting mass or be hinged when only being stressed when spring shaft is fixing or being hinged of compression spring end circle and supporting mass.

4. according to the described wheel of one of claim 1～3, it is characterized in that comprising the universal spring of described wheel rim (4) spring (3), one of the following non-inflatable tyre of connecting: one, band tyre; Two, poly-urethane rubber band tyre, there is cavity the centre; Three, polyurethane foam expanded cellular rubber obturator.

5. according to the described wheel of one of claim 1～3, it is characterized in that comprising the universal spring of described wheel rim (4) spring (3), the series connection air-inflation tyre.

6. according to the described wheel of one of claim 1～3, it is characterized in that comprising the universal spring of described wheel rim (4) spring (3), serial or parallel connection has one of following tire of air-inflation tyre shape, and 1, air-inflation tyre, bracing ply (23 ') be the thickening carcass plies or, steel ring shell or endless belt; 2, non-inflatable tyre, its bracing ply (23 ') carcass (23) one of or two are rubber-types, cord type or hard-shell type.

7. according to the described wheel of one of claim 1～3, it is characterized in that comprising the universal spring of described wheel rim (4) spring (3), one of following air-inflation tyre in parallel, 1, radial; 2, oblique; 3, band bundle oblique.

8. wheel according to claim 5, it is characterized in that described air-inflation tyre is one of following: the tyre surface that, comprises an adhesion bracing ply (23 '), (23) split of bracing ply (23 ') carcass or one, carcass (23) is flat annular arrangement shell or several sebific duct that a soft strengthening course is done, and is full of or is not full of sizing material between adjacent sebific duct; Two, comprise tyre surface and the plurality of parallel sebific duct below it, be with or without cushion layer between the two, be full of or be not full of sizing material between adjacent sebific duct.

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