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How do you calculate the volume of an irrregular object? - Answers
Unless you can partition it into regular object, you cannot. You would have to measure it rather than calculate it.One method that can be used for insoluble and relatively small and dense objects that are also solid(not containing hollow parts), is the method of displacement. Partially fill a large graduated container with a fluid (such as water). Record the volume of water, V1. Then gently insert the object into the fluid. Make sure it is fully submerged and then measure the volume of the fluid, V2. The volume of the object is equal to the difference in the two volumes, V2 - V1.This method is does not work in a wide range of real-life situations. It will fail with soluble objects unless you can find a fluid in which it is not soluble. With an unknown object you may not know what fluids will dissolve it.Obviously it will not work with objects, such as a large boulder, that are too big for any container. Or any object that cannot be moved, such as a statue on a plinth (unless you are prepared to spend time in prison!).The object needs to be wholly submerged and so measuring the volume of an object that is less dense than the fluid is not easy. You could get around this problem by attaching it to a denser object whose volume you know.Finally, it the object is hollow you will include the volumes of the cavities in your measure of the volume of the object.
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How do you calculate the volume of an irrregular object? - Answers
Unless you can partition it into regular object, you cannot. You would have to measure it rather than calculate it.One method that can be used for insoluble and relatively small and dense objects that are also solid(not containing hollow parts), is the method of displacement. Partially fill a large graduated container with a fluid (such as water). Record the volume of water, V1. Then gently insert the object into the fluid. Make sure it is fully submerged and then measure the volume of the fluid, V2. The volume of the object is equal to the difference in the two volumes, V2 - V1.This method is does not work in a wide range of real-life situations. It will fail with soluble objects unless you can find a fluid in which it is not soluble. With an unknown object you may not know what fluids will dissolve it.Obviously it will not work with objects, such as a large boulder, that are too big for any container. Or any object that cannot be moved, such as a statue on a plinth (unless you are prepared to spend time in prison!).The object needs to be wholly submerged and so measuring the volume of an object that is less dense than the fluid is not easy. You could get around this problem by attaching it to a denser object whose volume you know.Finally, it the object is hollow you will include the volumes of the cavities in your measure of the volume of the object.
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How do you calculate the volume of an irrregular object? - Answers
Unless you can partition it into regular object, you cannot. You would have to measure it rather than calculate it.One method that can be used for insoluble and relatively small and dense objects that are also solid(not containing hollow parts), is the method of displacement. Partially fill a large graduated container with a fluid (such as water). Record the volume of water, V1. Then gently insert the object into the fluid. Make sure it is fully submerged and then measure the volume of the fluid, V2. The volume of the object is equal to the difference in the two volumes, V2 - V1.This method is does not work in a wide range of real-life situations. It will fail with soluble objects unless you can find a fluid in which it is not soluble. With an unknown object you may not know what fluids will dissolve it.Obviously it will not work with objects, such as a large boulder, that are too big for any container. Or any object that cannot be moved, such as a statue on a plinth (unless you are prepared to spend time in prison!).The object needs to be wholly submerged and so measuring the volume of an object that is less dense than the fluid is not easy. You could get around this problem by attaching it to a denser object whose volume you know.Finally, it the object is hollow you will include the volumes of the cavities in your measure of the volume of the object.
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