Calculating the Radius of a Graduated Circle

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SUMMARY

The discussion centers on calculating the radius of a graduated circle using the arc-length formula, specifically for a problem from Loney's Trigonometry (1895). The key parameters are the division value of 5' (5 minutes of arc) and the distance between graduations of 0.1 inches. To find the radius, the formula \(s = r\theta\) is applied, where \(s\) is the arc length and \(\theta\) is the angle in radians. The confusion regarding the measurement of 5' being interpreted as either inches or seconds is clarified, confirming it refers to minutes of arc.

PREREQUISITES
  • Understanding of arc-length formula \(s = r\theta\)
  • Knowledge of angular measurements, specifically minutes of arc
  • Familiarity with basic trigonometry concepts
  • Experience with surveying instruments and graduated circles
NEXT STEPS
  • Research the application of the arc-length formula in surveying
  • Explore the historical context and usage of graduated circles in surveying
  • Learn about angular measurements and conversions between degrees, minutes, and seconds
  • Investigate modern surveying tools that utilize graduated circles
USEFUL FOR

This discussion is beneficial for students of trigonometry, surveying professionals, and anyone interested in the historical applications of geometric principles in practical tools.

Ragnarok7
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Hello, I was working problems from a very old trigonometry book, Loney's Trigonometry from 1895. There appears here a problem stating:

The value of the divisions on the outer rim of a graduated circle is 5' and the distance between successive graduations is .1 inch. Find the radius of the circle.

I cannot determine what is meant exactly by a graduated circle, other than that it was a surveying instrument. I am also unsure if the 5' measurement is supposed to mean 5 inches or 5 seconds (I'm thinking the latter). Does anyone have any ideas? Thank you!
 
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The expression 5' refers to 5 minutes of arc, or 1/12 of a degree (since there are 60 minutes of arc in a degree). So, I would use the arc-length formula:

$$s=r\theta$$

You are given $s$ and $\theta$, so you can solve for $r$. :D
 
Ah! Okay. I should have said minutes, not seconds. I get it now. Not sure why I was confused.

Thanks!
 

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