{"id":1578,"date":"2024-05-05T23:40:56","date_gmt":"2024-05-06T04:40:56","guid":{"rendered":"https:\/\/charles-oneill.com\/blog\/?p=1578"},"modified":"2024-06-09T22:04:38","modified_gmt":"2024-06-10T03:04:38","slug":"rotationally-symmetrical-tank-volume","status":"publish","type":"post","link":"https:\/\/charles-oneill.com\/blog\/rotationally-symmetrical-tank-volume\/","title":{"rendered":"Rotationally Symmetrical Tank Volume"},"content":{"rendered":"\n

This note demonstrates how to quickly find the volume of a rotationally symmetrical enclosure. I find this approach useful for estimating and designing tank volumes. As an example, you can quickly find the volume of a toroidal tank with a 1 inch circle revolved about a line 2 inch from the circle’s center is slightly under 10 cubic inches (precisely \u03c0 squared). <\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n

The equation is simple: Volume = 2\u03c0\u00b7A\u00b7L<\/em><\/strong>, where A is the area of the revolved cross section and L is the distance from the rotational axis to the center-of-gravity of the cross sectional area. For the torus, the area is 2\u03c0\u00b7\u03c0D2<\/sup>\/4\u00b7L which simplifies to \u03c02<\/sup>D2<\/sup>L\/2. Neat and fast.<\/p>\n\n\n\n

Here’s a design application: develop an equation for elliptical ended cylindrical tanks to hold precisely G gallons. Using the revolution equation in the figure below gives a total volume of V=\u03c0R2<\/sup>(h+4\/3\u00b7b). <\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n

Now conversion to gallons is a linear transformation with 231 in3<\/sup> per gallon. <\/p>\n\n\n\n

This approach is also useful for crazy, bizarre, strange, and otherwise normal volumes of revolution. For instance, you can now find the volume of a specialty triangular o-ring seal. <\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n

The triangular cross sectional area is half the base times height, so 50 mm2<\/sup>. The center of gravity of a triangle is 2\/3 the way from the tip to the base, so that’s 6.6 mm. The total distance from the axis to the CG is 20+6.6 = 26.6 mm. Multiply and you obtain: Volume = 2\u03c0 50 * 26.6 = 8377 mm3<\/sup><\/p>\n\n\n\n

Why does this work?<\/h2>\n\n\n\n

The key trick here is recognizing that in cylindrical coordinates, the differential volume is dV=r\u00b7dr\u00b7d\u03b8\u00b7dz, which appears to contain the moment equation dM=x\u00b7dx in a 2D Cartesian frame. A derivation is presented below.<\/p>\n\n\n\n

\"\"<\/a><\/figure>\n","protected":false},"excerpt":{"rendered":"

This note demonstrates how to quickly find the volume of a rotationally symmetrical enclosure. I find this approach useful for estimating and designing tank volumes. As an example, you can quickly find the volume of a toroidal tank with a 1 inch circle revolved about a line 2 inch from the circle’s center is slightly […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"_links":{"self":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts\/1578"}],"collection":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/comments?post=1578"}],"version-history":[{"count":6,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts\/1578\/revisions"}],"predecessor-version":[{"id":1596,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts\/1578\/revisions\/1596"}],"wp:attachment":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/media?parent=1578"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/categories?post=1578"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/tags?post=1578"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}