Could someone please explain how the value of sin is calculated when working out voltage drop for cables above 16mm? Do we use the values provided in BS 7671 appendix 4 such as resistance reactance and impedance of a conductor?
Thank you
Could someone please explain how the value of sin is calculated when working out voltage drop for cables above 16mm? Do we use the values provided in BS 7671 appendix 4 such as resistance reactance and impedance of a conductor?
Thank you
Please could you explain or highlight where you got this equation from
this one?
It's just a re-arrangement of the well known maths short-cut: sin²+cos² = 1
If you prefer, think of an right-angled triangle with a hypotenuse of 1. Pick one of the other angles, then sine of that angle give the length of the opposite side, cos of the same angle gives the length of the adjacent side. Then by Pythagoras - the square of the hypotenuse is equal to the sum of the square of the other two sides - so 1² = sin(angle)² + cos(angle)² - and 1² = 1 so sin(angle)² + cos(angle)² = 1.
- Andy.
Please could you explain or highlight where you got this equation from
this one?
It's just a re-arrangement of the well known maths short-cut: sin²+cos² = 1
If you prefer, think of an right-angled triangle with a hypotenuse of 1. Pick one of the other angles, then sine of that angle give the length of the opposite side, cos of the same angle gives the length of the adjacent side. Then by Pythagoras - the square of the hypotenuse is equal to the sum of the square of the other two sides - so 1² = sin(angle)² + cos(angle)² - and 1² = 1 so sin(angle)² + cos(angle)² = 1.
- Andy.