# Calculation of Rolling Impact of Helical Gear

The calculation of impact force in this paper is based on the principle of energy conservation at the moment of contact between the teeth.

The moment of inertia and induced mass of large and small wheels can be expressed as follows:

Ji=pi rhb2(rbi4_rhi4)(i=1,2)

Mi=Jirbi2(i=1,2)

Where Ji (i=1,2) is the moment of inertia of small and large wheels respectively, RH (i=1,2) is the density of gear material, B is the width of teeth, RBI (i=1,2) is the radius of the base circle of the large and small wheels respectively, and RHI (i=1,2) is the radius of the inner bore of the two wheel hubs respectively.

The impact kinetic energy of the gear pair at the rod impact point can be expressed as:

Ek=12m1m2m1+m2(v(12)) 2=12J1J2J1rb22+J2rb12(v(12)) 2

Where v(12) is the relative speed of large and small wheels at the rodent point

Due to impact, impact deformation Delta s occurs between the teeth, and the corresponding impact force Fs is the maximum impact force.Refer to Timoshenko’s assumption in elasticity [16]: if the equation between force and deformation established under static conditions is also established during the collision process, then the maximum impact force Fs can be obtained by equation (15).In the references, Timoshenko has studied the collision between two spheres. Based on the above assumptions, the maximum pressure and the collision duration between the two spheres during the collision process have been calculated. The theoretical calculation results about the relationship between the collision duration and the spherical radius have been verified by experiments by researchers.

FsFs=Ksδsn

In this paper, the coefficient n of the equation is obtained by fitting the data, which is taken as 1.1.

The elastic potential energy is expressed as:

Ek=0delta sKsxndx=1n+1Ksdelta sn+1

According to the theory of impact mechanics, there is a relationship between impact kinetic energy and elastic potential energy as follows:

Ek=12J1J2J1rb22+J2rb12(v(12)) 2=1n+1Ks Delta sn+1

Therefore, the maximum elastic deformation is expressed as:

Deltas=(n+12J1J2(J1rb22+J2rb12)ks(v(12))2)

Finally, based on the maximum elastic deformation obtained, the maximum impact force can be expressed as (the impact force is calculated without considering the effect of friction):

Fs=(n+12J1J2(J1rb22+J2rb12) (v(12)) 2Ks1n)n n+1

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