No, this has nothing to do with the American economy. There is another difficulty with the calculation that Rashi uses to conclude that the

*teiva*was submerged 11*amos*. How could Rashi base his calculation on the depth of the water decreasing at a constant rate. One can generally assume that when water decreases, it does so at a constant rate of volume. However, mathematically, if the volume of a sphere decreases at a constant rate, the rate of change of the depth will increase as the waters become shallower. The shallower the water gets, the faster it will decrease depth-wise. How then could Rashi assume that the depth decreased at a constant rate? This is the question posed by מהרי"ל דיסקין. He gives his own answers to this question. One, for instance, is that the waters receded, the ground became more saturated which slowed down the overall receding process and hence balanced out the constant rate of change of depth. But a*rebbe*of mine from Yeshivas Ohr Yerushalayim posed this question of none other than Nobel Prize winner Yisrael Aumann. He answered simply that mathematically, none of this is needed. True, the rate of change of depth is not directly proportional to the rate of change of volume. However, considering the size of the globe, the difference between the two within the scope with which we are dealing is negligible and would not affect Rashi's calculation. Is this true? The short answer is "Yes". The longer answer requires a little Calculus.
The radius of Earth is 6372500 m. To make things simple we will convert this to

*amos.*Instead of using three separate measures of the*amah*, we will keep things neat and use an average figure of 53 cm. (6372500 ÷ 0.53 = 12023585) That translates to 12023585*amos*. To make things simpler, we will round it off to 12000000*amos*. This will have little effect on the final outcome. This figure will be called r_{w}.
The standard equation for volume:

V

_{w}= 4/3πr_{w}^{3}
Through implicit differentiation:

ΔV= 4πr

_{w}^{2}Δr,
where ΔV is the rate of change of volume and Δr is the rate of change of radius. We have already set r

_{w}to be 12000000 and Δr is ¼ (*amos*/day according to_{}Rashi). Therefore,
ΔV= 4π (12000000)

^{2}¼
ΔV= 4.524 x 10

^{14}(constant)
The goal of these calculations is to see whether or not Δr changes significantly over the course of the decreasing of the water. To see how much Δr changes, we must switch around the equation to define Δr and instead of using the figure of 12000000 for the radius, we will use the new radius when the top of the mountains became visible, 11999985.

As stated before, ΔV= 4πr

_{}^{2}Δr
Therefore, Δr

_{2}= ΔV/ 4πr_{new}^{2}
Δr

_{2}= 4.524 x 10^{14}/ 4π(11999985)^{2}
Δr

_{2}= 0.2500006250012
This means that if the waters were receding at a rate of change of depth of 0.25

*amos*per day when they began receding, then 60 days later they were receding only 0.0000006250012*amos*/day faster, a rather negligible amount indeed.
I think this is a cool vort, but since i cant understand it, I cant say that it is.

ReplyDeleteHere I am again, nearly three and a half years later, returning to read this vort, having been led here via a link from the weekly shtikle. And once again, all I can say was already said in previous comment: I'm not smart enough to understand this, so I can't say for sure whether or not this is a brilliant vort. Sure LOOKS brilliant, though.

ReplyDeleteIf 12,000,000 is the radius of the earth, how can you say that it's the radius of the earth plus the waters on top of it?

ReplyDeleteI'm simply using a round number as a starting point. Since my starting point is the radius including the water, that is what the 12,000,000 represents.

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