{"id":4012,"date":"2010-01-06T22:02:35","date_gmt":"2010-01-06T14:02:35","guid":{"rendered":"http:\/\/www.intmath.com\/blog\/?p=4012"},"modified":"2014-10-29T18:20:12","modified_gmt":"2014-10-29T10:20:12","slug":"arc-length-for-the-inner-curve-of-a-window","status":"publish","type":"post","link":"https:\/\/www.intmath.com\/blog\/mathematics\/arc-length-for-the-inner-curve-of-a-window-4012","title":{"rendered":"Arc length for the inner curve of a window"},"content":{"rendered":"

A reader who works for a glass company (true story) wrote to me recently and asked how to solve the following. <\/p>\n

\n

I've got to make a window with a curved top. The width of the frame is the same all round, including the part around the curved portion. <\/p>\n

What's a formula for the length of the inner arc of the curved portion?\n<\/p>\n<\/blockquote>\n

The required length is labeled HI in the diagram.<\/p>\n

\"window<\/div>\n

(This is not quite a \"Norman window\", since it's not a semi-circle on top.)<\/p>\n

This is a typical \"real life\" question, in that we don't have a lot of information to go on, so we'll need to make some assumptions.<\/p>\n

Solution - Example<\/h2>\n

Let's consider a plausible example first. We assume the curves are arcs of circles. (They looked circular in the question. If they are not circles we could adjust later, but for the sake of the example, we'll stick to this reasonable assumption.)<\/p>\n

Let the total frame (length CD) be say 4 units wide and the edges of the frame be 0.3 units wide. I pick a point P in the center (horizontally) of the frame (I choose point P (2,1)), and draw 2 concentric circles, 0.3 units apart. The outer one is 5 units, the inner one 4.7 units. I could have chosen any radii for my concentric circles, of course, as long as the inner and outer radii differ by 0.3 units.<\/p>\n

\"window<\/div>\n

If I can find angle θ<\/span>, it will be straightforward to find the arc length HI.<\/p>\n

First, we find angle α<\/span> by using the right triangle PMI.<\/p>\n

cos α = 1.7 \/ 4.7 = 0.3617<\/span><\/p>\n

Using the inverse ratio, we get that <\/p>\n

α = arccos (0.3617) = 1.2007<\/span> (radians, of course. If we need degrees, it equals 68.79°)<\/p>\n

[Why radians? They are more commonly used in science and engineering than degrees, and are best for this problem. For more, see Radians<\/a>.]<\/p>\n

Now we observe that 2α + θ = π (180°)<\/span>, since they lie on a straight line.<\/p>\n

So angle θ = π - 2 × 1.2007 = 0.7402<\/span><\/p>\n

To find the arc length HI, we just apply the arc length formula<\/p>\n

s = r<\/em> θ<\/span><\/p>\n

(See arc length formula<\/a>)<\/p>\n

s<\/em> = 4.7 × 0.7402<\/span><\/p>\n

= 3.4789<\/span><\/p>\n

So the required arc length is 3.48 units (correct to 2 decimal places).<\/p>\n

General Solution<\/h2>\n

Let the total width (window and frame) = 2x<\/em><\/span>, giving x<\/em><\/span> for half the width.<\/p>\n

Let the edges of the frame have width w<\/em><\/span>.<\/p>\n

The circles have radius R<\/em> (outer) and r = R - w<\/em> (inner).<\/p>\n

Angle α = arccos ((x - w<\/em>) \/ r<\/em>)<\/span><\/p>\n

Angle θ = π - 2 α<\/span> (in radians)<\/p>\n

Arclength HI is given by<\/p>\n

s = r<\/em> θ<\/span><\/p>\n

= r<\/em> × (π - 2 (arccos ((x - w<\/em>) \/r<\/em>)))<\/span><\/p>\n

This is the formula required by the glass manufacturer.<\/p>\n

See the 13 Comments<\/a> below.<\/p>\n","protected":false},"excerpt":{"rendered":"

\"you<\/a>A glass manufacturer asked me how to find the length of the inner arc of a circular window frame.<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mo_disable_npp":""},"categories":[4],"tags":[],"_links":{"self":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts\/4012"}],"collection":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/comments?post=4012"}],"version-history":[{"count":0,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts\/4012\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/media?parent=4012"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/categories?post=4012"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/tags?post=4012"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}