You need to remember that the volume of gas is influenced by temperature and pressure and that gases expand to fill any container in which they're placed. We have to use more elaborate methods to measure the volume of a gas. So, if you want to measure an irregular object's volume, just follow in Archimedes' footsteps (though you can omit the naked race part): Legend says that Archimedes was so excited about this discovery that he popped out of his bathtub and ran naked through the streets of Syracuse. Knowing the irregular object's volume and its weight, he could calculate the density and compare it with the density of pure gold. From this observation, he deduced that volume of water displaced must be equal to the volume of the part of his body he had submerged. The idea came to him when he was taking a bath - stepping into a bathtub, he noticed that the water level rose. If it's an irregular shape, you can try to do the very thing that caused Archimedes to shout the famous word Eureka! Probably you heard that story - Archimedes was asked to find out if the Hiero's crown is made from pure gold or just gold-plated - but without bending or destroying it. For a right triangular prism, the equation can be easily derived, as well as for a right rectangular prism, which is apparently the same shape as a box.įor regular three-dimensional objects, you can easily calculate the volume by taking measurements of its dimensions and applying the appropriate volume equation. Prism = A h Ah A h, where A A A is a base area and h h h is the height. For a pyramid with a regular base, another equation may be used as well: Pyramid = ( n / 12 ) h s 2 cot ( π / n ) (n/12) h s^2 \cot(\pi/n) ( n /12 ) h s 2 cot ( π / n ), where n n n is a number of sides s s s of the base for a regular polygon. Pyramid = ( 1 / 3 ) A h (1/3)Ah ( 1/3 ) A h where A A A is a base area and h h h is the height. Rectangular solid (volume of a box) = l w h lwh lw h, where l l l is the length, w w w is the width and h h h is the height (a simple pool may serve as an example of such shape).
Sphere = ( 4 / 3 ) π r 3 (4/3)\pi r^3 ( 4/3 ) π r 3, where r r r is the radius.Ĭylinder = π r 2 h \pi r^2h π r 2 h, where r r r is the radius and h h h is the height.Ĭone = ( 1 / 3 ) π r 2 h (1/3)\pi r^2h ( 1/3 ) π r 2 h, where r r r is the radius and h h h is the height. Here are the formulas for some of the most common shapes:Ĭube = s 3 s^3 s 3, where s s s is the length of the side. There is no simple answer to this question, as it depends on the shape of the object in question.
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