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VE1AJF > TECH 21.07.26 05:02l 40 Lines 1656 Bytes #999 (0) @ WW
BID : 37089_VE1MPF
Read: OE5RCO GUEST DJ6UX
Subj: Quick Calc 5-stud wheel circle.
Path: DB0FHN<OE2XZR<OE6XPE<DB0RKB<DK0WUE<HB9ON<IK7NXU<IK6IHL<IZ3LSV<IR1UAW<
IR1UAW<IR0AAB<VE3CGR<VE3QBZ<VA3AMJ<VE1MPF
Sent: 260721/0246Z 37089@VE1MPF.#MCTN.NB.CAN.NOAM BPQ6.0.24
To find the bolt circle diameter of a 5-stud automotive (or other) wheel:
- Measure (in inches), O.C. between any 2 "wide-spaced" studs/holes.
We'll refer to this dimension as "Y".
- Diameter of Bolt Circle (in inches) = 1.0514 x Y
*****************************************************
Above was derived using trigonometry with the dimension "Y"
being taken from any 2 of the wide-spaced studs/holes to allow for
best accuracy. (any 2 adjacent holes could have been used but the
measurement accuracy is more critical).
Derivation: 5-bolt wheel holes/studs are 72 deg apart on bolt circle.
- Measure O.C. direct distance between any hole/stud and a stud/hole
2 away (or 144 degrees) from it. This is distance "Y" referred to above. Divide
Distance "Y" by 2...this becomes the "Opposite" side of a right-
angle triangle, used to calculate it's "Hypotenuse" (which is the
bolt-circle radius we'll call "r"). Ultimately our desired bolt-circle
dimension is 2r
- Using Trig, where SineAngle = "opposite side" / "Hypotenuse side" where
Sine 72deg = .9511 we rearrange to solve for "r" with:
r = 0.5Y/.9511 and our bolt circle diameter is 2r.
ex: Inital cross-stud measurement "Y" was 4-9/32" or 4.28125"
0.5Y is half of this or 2.140625"
thus r = 2.140625/.9511 = 2.2507" and
D = 2r or 4.50" so our bolt pattern is 5 on 4.5" nominally.
- This equation was reduced so as to solve all common values to a constant of K
value = 1.0514 to make quick measurements easier.
73 - Gord VE1AJF
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