h a l f b a k e r yWhy not imagine it in a way that works?
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At the moment I do not have access to one of the many watt to time conversion charts on the internet.
I have an 800 W microwave. The frozen dinner I am preparing says to cook it at 1100 Watts for 6 1/2 minutes.
I punch the 1100 watts for 6 1/2 minutes buttons on the built-in microwave calculator
to calculate the additional time it will take to heat the food in a lesser 800 W microwave.
The microwave automatically sets the microwave for the new time as acquired through the calculator.
Edit 2020 3 21: I should have mentioned "Because there are so many variables when it comes to micro-cooking frozen foods, there is not a specific formula or direct ratio for calculating conversion times."[1]
No Algebra required at school [2]
People are bad at math [3]
[1] Maybe the proper conversion can only be done by the manufacturer of the microwave?
https://www.theeagl...4-e31a9377dfe0.html [Sunstone, Mar 21 2020]
[2] No algebra required at school to teach you to learn how to think
https://duckduckgo....uired&t=ffsb&ia=web [Sunstone, Mar 21 2020]
[3] Many if not most are bad at math
https://duckduckgo....+math&t=ffsb&ia=web [Sunstone, Mar 21 2020]
[link]
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6.5 minutes = 390 seconds. |
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390 * 1.375 = 536.5 = 8.93 minutes |
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No online calculator needed... |
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Perhaps the microwave oven control panel should be calibrated in whatt-ours instead of seconds? Would that be in joules I suppose? |
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It would be better to calculate a set of conversion factors - in binary, for universal applicability - for each power range; these could then be chiselled onto a small polished granite stele, placed alongside the microwave oven for easy reference. |
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We guess RFC 1149, "IP over Avian Carriers" |
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But that does exactly fulfill the criteria for half-bakedness ... |
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It would have to be a Babbage engine, powered by a Newcomen engine driven by steam from the microwave oven, then. .. |
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I think you should need to increase the time beyond what the formula says when scaling for a less-powerful microwave oven, and vice versa, because heat loss from the food will take away a larger portion of the input energy at a lower power, due to the heating taking place over a longer time interval. By how much will depend on the food and its shape. |
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