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In this question it is asked what the upper-bound of the ratio of a solid object's surface area can be visible through direct, unaided observation.

The accepted answer says that there is no such upper-bound, referring to a function whose mathematical limit is $1$ as the value $r \rightarrow 0$.

I am not sure how to interpret this. I see two interpretations; one that is correct but perhaps somewhat misleading* $(1)$, and one that is just plain illogical $(2)$.

$(1)$ There is no upper-bound, because the proportion grows arbitrarily close to $1$ and due to the density of the reals, there is no one number to grasp at. This is true.

$(2)$ The upper-bound is $1$, because the function is, mathematically, continuous over $r=0$. Its analogical description of physical reality however, ends at $r=0$. As such, viewing the function as a mathematical model of a physical transformation, the function is discontinuous at $r=0$, as an a priori truth following from the definition of a cone (it obviously can't be a cone if its radius is $0$).

If the answerer's intended interpretation was $(2)$, then this seems to me as a situation of someone conflating a mathematical analogy with the physical reality it is describing. Just because the math say that $r$ can be $0$, doesn't mean that physical reality allows $r=0$.

*The answer itself isn't really misleading, but the answerer's comments in the discussion confused me. John Deaton pointed to the fact that the answer did not explicitly rule out $r=0$ as a possibility. The answerer said that "the answer left that possibility untouched", and that they were "uninterested in that analysis". This means that despite the fact that $(1)$ is the only logical interpretation for the cone, the answerer explicitly stated that both $(1)$ and $(2)$ are possible interpretations. Their language (to me) within the answer seems to suggest $(1)$, but their comments in the discussion states either is possible. The most confusing part is that I find $r=0$ to be trivially illogical; but the answerer stating they were "uninterested in that analysis" suggest that from their point of view, it isn't trivial, but rather something to be analyzed. Given that they appear quite competent to me, this makes me doubt my feeling that $r=0$ is trivially illogical.

The question:

If $r=0$, there is no base, and thus, there is no cone. Given that this is a simple fact, why is the elimination of the case of $r=0$ referred to as an analysis, when it appears to be a fact trivially establishable? Is this as simple as I think it is, or is there more going on here?

Qmechanic
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user110391
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  • But the math doesn't allow for $r=0$. Check out the unapproximated equation. When $r=0$ it is undefined. Furthermore, the question isn't taking the limit as $r\to0$ but rather $r/h\ll1$. These are two different things. – BioPhysicist Apr 05 '22 at 01:34
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    Emilio's comments seem quite clear to me: your initial interpretation (1) is fine, and he is just uninterested in spending lots of time quibbling over the semantics of what happens "at" $r = 0$. That is an uninteresting degenerate case, like asking what the largest element of an empty set is, or whether a line segment counts as a rectangle, or what the angles are for a triangle where all three vertices coincide, or the number of ways to choose zero items from an empty list. It's all equally debatable and equally irrelevant. – knzhou Apr 05 '22 at 01:36
  • @BioPhysicist That's true, but his last expression isn't discontinuous over $0$. That isn't a problem for me, but it did spawn a discussion. In that discussion, this degenerate case of $r/h$ was brought up. The conclusion of that discussion offered by Emilio was that he was simply uninterested in analyzing that case. As I said, I didn't find his answer misleading, I found his comments misleading. It would have been a different thing if he simply said "my answer doesn't leave that option open", which is arguable, or if he said "that case is so nonsensical I didn't bother touching on it." – user110391 Apr 05 '22 at 01:42
  • @knzhou He spent more time quibbling over completely unimportant when a simple "the function is obviously discontinuous over $r=0$ since a cone must have a radius", which he could have followed up with either "I thought my answer showed my model didn't allow for $r=0$", or with a "I found it so obviosly nonsensical that touching on it was unnecessary". Instead however, it seemed like his comments were saying that "r=0" was a possibly possible case he hadn't bothered to analyze. That's what I found weird. – user110391 Apr 05 '22 at 01:46
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    I'm voting to close this question because it seems to me that the only person able to provide a non-opinion based answer would be Emilio, since the question is concerned with the meaning of the choice of words made in his answer (from the comments in this post, it seems everyone agrees with the Physics and Math of the problem) – Níckolas Alves Apr 05 '22 at 02:24
  • @NíckolasAlves Then you have misunderstood the question. I'm not asking about what Emilio meant, I'm using his comments as an explanation for my confusion about X, and then I ask about X. What is X? Whether or not the function being discontinuous over $r=0$ is a trivial fact or not. That is my question. I have established my understanding, the cause of my doubt of that understanding, and the question that I'm left with. An answer to that does not require (though would perhaps be helped by) Emilio's explanation. So, is this is as simple as "the function is obviously discontinuous over $r=0$"? – user110391 Apr 05 '22 at 03:00
  • @user110391 Isn't this a mathematics question then if you are just asking if a function is discontinuous or not? Obviously the actual ratio that does not have an approximation is undefined there. The approximation is defined there only because it is using essentially a linear approximation in $r$. I don't see how your accepted answer answers the question of continuity, but if it helped answer the question for you then great :) – BioPhysicist Apr 05 '22 at 03:18
  • @BioPhysicist I posed two possibilities: either my simple understanding that the function is obviously discontinuous over $r=0$ is correct, or there is something more going on. The accepted answer simply states "the function is discontinuous over $r=0$", which I take as stating "no, there isn't something more going on". Assuming this quite non-radical stance is correct, I'm left to conclude that Emilio probably didn't bother to include the "$r>0,h>0$" domain limitation in his answer (given that it is kinda obvious), and that he perhaps was being sarcastic with his word-choice of "analysis". – user110391 Apr 05 '22 at 03:25
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    General tip: Referring to the "accepted answer" is ambiguous at it may change in the future. – Qmechanic Apr 05 '22 at 07:28

1 Answers1

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The equation being referenced is

$$ R = 1 - \frac r h , \qquad r>0, h>0$$

An upper bound on the ratio $R$ is $1$ because $1$ is equal to or greater than any value of $R$ which can be achieved from any valid choice of $r,h$. It also happens that $1$ is the smallest upper bound or supremum.

Myridium
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