The shortest effective cutting length of a gear hob is the minimum axial length (L0t) min of the hob required to cut out the entire tooth height of the gear. In the machining of gears, there are many cases in which the shortest effective cutting length of the hob needs to be calculated. If the pinion gear of the double gear is hobbed, if the pinion gear is a helical gear and is close to the large gear, it is necessary to check whether the hob hobbing pinion will collide with the large gear. The smaller the outside diameter of the hob, the shorter the axial length and the less likely it is to collide with the bull gear. However, if the outer diameter of the hob is too small, it will affect the strength of the tooth root slot (in this case, the hob may be considered to be integral with the shaft); in addition, if the axial length of the hob is too short, it may not be able to cut out the entire small piece. gear. Therefore, it is necessary to calculate the shortest axial length of the hob in this case. In addition, (L0t)min is also the basis for calculating the total length of the hob tooth (taking into account the length of the string). 1 Calculation method According to Fig.1, (L0t)min can be derived when the roll spur gear is calculated. For ordinary precision hobs, since the helix angle is very small, the normal tooth angle can be considered equal to the axial tooth angle. The intersection of the meshing line passing through point P with the diameter of the gear tip circle is a, and the intersection point with the hob tooth top line is b, then the workpiece tooth profile is formed between the meshing lines a and b. The perpendicular line passing the a point and the top line of the hob tooth intersect at c', and the crest angle at the outer side of the c' cutter is c, the shortest distance of the hob tooth part along the normal direction (ie no string is required). The knife can cut out the shortest length of the gear's total tooth height.) (L0n) min = 2 (pm/2+CE) = pm + 2L3 (1) Figure 1 Calculate the half-to-half pitch of the hob. Pm/2), because it is assumed that the tip of the hob is higher than its root. Although it is not actually required that the height of the hob tooth be equal to its root height, it is only required that the full height of the hob tooth is greater than the full height of the workpiece tooth, and that the height of the hob tooth is higher than the height of the workpiece root (the height of the workpiece tooth root is (fa1+) C1) m, where fa1 is the height coefficient of the workpiece tooth top, c1 is the radial clearance factor, and the tooth root position is shown in RS in Figure 1), but it is more troublesome to substitute the high RS of the workpiece root. To simplify the calculation, it may be assumed that the root of the hob is higher than the height of the hob tooth, at this time RS = pm/2. Because the root height of the actual hob is usually lower than the workpiece root height, the value of the hob (L0n) min calculated according to formula (1) is slightly increased, and the calculation result is more secure. It can be seen from FIG. 1 that in order for the hob to cut out the complete tooth shape, L3≥L2 must be satisfied, because the shortest length of the hob needs to be calculated, so its extreme condition L3=L2 can be taken. L2 is the actual length involved in the cutting. This length is taken when machining the double gears for calculation. The formula is L2 = L1 + Qa = L1 + QC' tana1 = L1 + EN tana1 = L1 + (NP + ME-PM) Tana1 = L1 + [L1tana1 + (fa1 + c1) m - x1 m] tana1 = (tan2a1 + 1) L1 + (fa1 + c1 - x1) mtana1 (2) where: L1 = Ra1sin (aa1 - a1) Ra1 - tooth of the workpiece The top circle radius aa1 - the workpiece top circle pressure angle, cosaa1 = rb1/Ra1 = mz1cosaa1/ Ra1 After obtaining L2, take L3 ≥ L2, substituting (1) to find (L0n) min, hob The shortest axial length is (L0t)min = (L0n)min/cosg0 (3) where g0 is the helix angle of the hob because g0 is usually small (about 3°, cos3° ≈ 0.998), whereas The shortest effective cutting length of the hob is finally rounded to mm, so (L0n)min can also be directly used as (L0t)min. Although the above formula is derived from the case of roller spur gears, if the relevant parameters are replaced with normal parameters, equivalent gears are used instead of workpieces, and they can be similarly used for the calculation of helical gears. If you need to check whether the pinion will collide with the large gear when the double gear is used to roll, double the L2 calculated by equation (2) as (L0t) min, ie (L0t) min = 2L2 is the roll that participates in the cutting The shortest length of knife. The value calculated by equation (1) is the shortest length that the hob should have when it is not stringed. 2 Application Examples Example 1: The parameters of our factory's gear workpieces are: da1=96.2mm, radial pitch P=8 (ie m=3.175mm), z1=29, a1=20°, tooth full height h1=5.733mm, tooth Top height ha = 2.06mm. The gear was hobbed by a single-head hob with an outer diameter of 90 mm and a pitch circle diameter D0 = 82.654 mm, and the shortest length of the hob tooth was found. Solution: First calculate the workpiece tip circle pressure angle aa1: cosaa1=mz1cosaa1/Ra1=0.8993991382
Aa1=25.9208°
L1=Ra1sin(aa1-a1)=4.96mm The workpiece root height coefficient is fa1+c1=(h1-ha1)/m=1.15
L2=(tan2a1+1)L1+(fa1+c1-x1)mtana1=6.94mm Since L3>L2, L3=7 is assumed. The shortest length of the hob is (L0n)min=pm+2L3≈24mm. The hob is single-headed and its helix angle is g0=arcsin(m/D0)=2.201452653 then (L0t)min=(L0n)min /cosg0=24.01mm It can be seen that the error caused by (L0n)min as (L0t)min is negligible. From the above calculations, it can be seen that L1 = Ra1 sin (aa1 - a1) has a greater effect on the value of (L0n)min, and Ra1 and aa1 have a greater effect on L1. The larger the workpiece, the larger the Ra1 and aa1, and the longer the (L0n)min of the reel. In addition, from equation (2), it can be seen that the larger the tooth root height of the workpiece, the smaller the coefficient of displacement (in this example, x1=0), the larger the (L0n)min of the hob. Example 2: Helical gear mn = 3.5mm, an = 20°, z1 = 100, b1 = 60°, h1 = 7.875mm, ha1 = 3.5mm (for analyzing the effect of each parameter pair on (L0n)min, for b1, etc. The parameters are slightly exaggerated). Let the workpiece be a double gear, and check whether or not it collides with the large gear during the machining process and calculate the (L0t) min of the gear hob. Solution: The equivalent number of teeth of the helical gear is z1'=z1/cos3b1=181.93≈182 The equivalent diameter of the gear is mnz1'=637mm, and the diameter of the top circle is mnz1'+2ha1=644mm. At this time aa1=arccos[mnz1'cosan (/mn z1'+2ha1)]
= 21.64632097° L1 = Ra1'sin(aa1-a1)≈9.25mm fa1+c1=(h1-ha1)/mn=1.25 L2=(tan2an+1)L1+(fa1+c1-x1)mntanan
=12.067mm The minimum length of the hob (L0t)min=2L2≈24mm. At present, the hob design method commonly used in China is to determine the minimum length of the hob by listing the maximum value of the relevant coordinates of the elliptic equation of the hob oblique section. Because the elliptic equation is involved, the derivative is more troublesome. In order to find the maximum value of the coordinate, the solution beyond the equation is also required, which is very difficult for manual calculation. Calculating (L0t)min using the largest coordinate value is rough. In contrast, the method described in this paper is simpler and more practical, and the calculation results of the two methods are relatively close (L0t)min≈24mm obtained in Example 2 is slightly larger than (L0t)min=22.7mm obtained by the ellipse method. Hobs are safer and more reliable when used.
Aa1=25.9208°
L1=Ra1sin(aa1-a1)=4.96mm The workpiece root height coefficient is fa1+c1=(h1-ha1)/m=1.15
L2=(tan2a1+1)L1+(fa1+c1-x1)mtana1=6.94mm Since L3>L2, L3=7 is assumed. The shortest length of the hob is (L0n)min=pm+2L3≈24mm. The hob is single-headed and its helix angle is g0=arcsin(m/D0)=2.201452653 then (L0t)min=(L0n)min /cosg0=24.01mm It can be seen that the error caused by (L0n)min as (L0t)min is negligible. From the above calculations, it can be seen that L1 = Ra1 sin (aa1 - a1) has a greater effect on the value of (L0n)min, and Ra1 and aa1 have a greater effect on L1. The larger the workpiece, the larger the Ra1 and aa1, and the longer the (L0n)min of the reel. In addition, from equation (2), it can be seen that the larger the tooth root height of the workpiece, the smaller the coefficient of displacement (in this example, x1=0), the larger the (L0n)min of the hob. Example 2: Helical gear mn = 3.5mm, an = 20°, z1 = 100, b1 = 60°, h1 = 7.875mm, ha1 = 3.5mm (for analyzing the effect of each parameter pair on (L0n)min, for b1, etc. The parameters are slightly exaggerated). Let the workpiece be a double gear, and check whether or not it collides with the large gear during the machining process and calculate the (L0t) min of the gear hob. Solution: The equivalent number of teeth of the helical gear is z1'=z1/cos3b1=181.93≈182 The equivalent diameter of the gear is mnz1'=637mm, and the diameter of the top circle is mnz1'+2ha1=644mm. At this time aa1=arccos[mnz1'cosan (/mn z1'+2ha1)]
= 21.64632097° L1 = Ra1'sin(aa1-a1)≈9.25mm fa1+c1=(h1-ha1)/mn=1.25 L2=(tan2an+1)L1+(fa1+c1-x1)mntanan
=12.067mm The minimum length of the hob (L0t)min=2L2≈24mm. At present, the hob design method commonly used in China is to determine the minimum length of the hob by listing the maximum value of the relevant coordinates of the elliptic equation of the hob oblique section. Because the elliptic equation is involved, the derivative is more troublesome. In order to find the maximum value of the coordinate, the solution beyond the equation is also required, which is very difficult for manual calculation. Calculating (L0t)min using the largest coordinate value is rough. In contrast, the method described in this paper is simpler and more practical, and the calculation results of the two methods are relatively close (L0t)min≈24mm obtained in Example 2 is slightly larger than (L0t)min=22.7mm obtained by the ellipse method. Hobs are safer and more reliable when used.
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