Math Is Fun Forum

Hi Bobby;

The minimum slider increment is 0.00000001, so a slider on C can't get the accuracy. Pity the minimum can't be set closer to the max rounding. But at that level of accuracy, the length of the slider would have to be increased greatly to get the desired effect...you'd have to be zoomed right in before you could use it at that extreme level.

Anyway, I can now get Gebra's maximum (I think) string height accuracy of 3888.61446059 (for which B3 = 99.99999999988358) with my zooming routine in about 2 minutes, including using zoomin[1000] and centreview[C] three times at the end.

Gebra gives four more decimal points after the 9 (ie, 4952), but I've not been able to get it to show the next digit, 3, that is given by W|A, which has 3888.614460593728 (correct to 12 decimal places) as its result, working at 100-digit accuracy.

It's the second tool in the third-from-the-left toolbox in the Toolbar at the top of the page, in the same toolbox where the Line tool lives. I say "Line tool" hesitantly, which you'll understand if you read on...

On the Gebra site:

Why do that? Clever!

I wonder what it's called in English-speaking countries other than Aus, UK & US?

Hi thickhead;

Well, that's very interesting. That's pretty much my post #21's result, where I'd used M's FindRoot to get 0.01928719490295426 (ie, radians...my formula was in degrees), resulting in 3888.614460591227.

By increasing working precision in my M to 100 digits, I now get 0.019287194902954844824414790282500 and 3888.61446059372823946337916760 (just showing the first few decimal places for both).

W|A gives the same result as above for theta with 41807040Tan[th]-100-41807040th=0 (41807040 being 2*3959*5280), and, after plugging in 100-digit theta, also gives the same result as above for the string height with 20903520*Sec[th] - 20903520 (20903520 being 3959*5280).

Sorry, I don't understand your series approach...my maths never got that far.

In Geogebra I've not been able to work out how to code exactly 100 feet for the extra bit of string in order to get Geogebra's most accurate value for the string height above the earth. By grabbing the string vertex and manually dragging it to &/or from the earth, I've got as close to 100 as 99.9999999984866 (by zooming right in on the vertex and adjusting it), which gives 3888.6144606396556. And that's near enough to close enough for me!

Plugging 3888.61446059372823946337916760 into Geogebra gives 99.99999999988358. Your 3888.614461, which looks like being either truncated or rounded, gives a slightly less accurate result.

I think our results show that my Geogebra method is good, so that gives me more confidence in doing as Bobby has suggested, which is to write it up in the Computer Math forum.

So, thanks for reporting your latest finding.

When I saw you leap into this thread I thought you were out to surprise me!

Without Geebra I'd've been nowhere on this problem...and using it was a pleasure.

M too.

Hi Bobby;

Sorry...my mistake. I'd read into what you'd been writing that you were saying that the second answer was the right answer to the problem.

Correct!

It'll never happen to me, though.

...so now you know when my birthday isn't.

When I plug Bobby's W|A result (from post #20's formula) into G - which I hadn't tried before - I get 3889.00031521 (correct to 8 decimal places), so my G method seems to be good.

The error looks like being the central angle, for which I get different answers depending on whether I use Bobby's formula or the surveyor site's.

I've probably messed up the surveyor formula somehow...

Hi thickhead;

Thanks for your answer - and excellent thought - but single-digit ages are not exactly what I had in mind. Maybe my wording's a bit loose.

It's probably ok mathematically on occasion, but not really when applied to ages. That's here, down under, in Australia, anyway...but maybe we're a little different from the rest of you.

When their great day arrives, a 07-year-old kid is hardly likely to say to their classmates, "Hey! Guess what! I'm zero-eight today!!" Then they'll have zero-one heck of a job trying to explain to the other single-digit-year-olds in their class what on earth they meant by that! It'd probably take the kid until well after their 05-past-03 home time before any-zero-one would get it!

Please forgive me!

Maybe that's why my answer doesn't make sense to me, as I said earlier. I'd expected the correct answer to be closer to the height of an atom (Mathgocart's answer a).

According to an online calculator into which I entered the earth's diameter (or radius?) and some other figure I don't recall, 3890 feet also happens to be the perpendicular distance from the top of the earth down to the chord between the two tangent points. I thought that was rather coincidental, and so I started to doubt my understanding of my version of the problem...but the huge scale isn't helping me to get my head around it.

Sorry...have to go out for 5 hours or so.

Hi all;

I understand (maybe wrongly) from Mathegocart's explanation that the shape for the longer string version is circular (like the original string), but in an experiment in which I used my basketball the shape is quite different (see image).

The string curves around the ball tightly from the base to the tangent points (the black dots), and then both sides continue as straight lines up to the suspension point...the apex.

Transferring that shape to Mathegocart's problem I've tried to work out the distance from the top of the earth to the string's apex, using the popular integer measurement of 40075km (that I converted to imperial) for the earth's equatorial circumference and basing the diameter on that.

But...I'm getting an answer that doesn't make sense to me.

What answer do you guys get?

Thanks!