## Correction Factor

Consider a helical torsion spring is made from a 0.5 cm diameter wire. If the mean spring diameter is 50 mm, what is the correction factor?

Expand Hint
Correction factor:
$$K_i=\frac{(4C^2-C-1)}{[4C(C-1)]}$$$Hint 2 $$C=\frac{D}{d}$$$
where $$D$$ is the mean spring diameter and $$d$$ is the wire diameter.
Correction factor:
$$K_i=\frac{(4C^2-C-1)}{[4C(C-1)]}$$$where $$C=D/d$$ , $$D$$ is the mean spring diameter, and $$d$$ is the wire diameter. First, let’s solve for $$C$$ : $$C=\frac{50mm}{5mm}=10$$$
Thus,
$$K_i=\frac{(4(10)^2-(10)-1)}{[4(10)(10-1)]}=\frac{(400-10-1)}{[40(9)]}=\frac{389}{360}=1.08$$\$
1.08