"Rusty Vandura - www.tinyurl.com/keepoppo" (rustyvandura)
05/24/2019 at 11:30 • Filed to: None | 1
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[EDIT]: Thank you.
A 16-foot chord is inscribed on a circle with a radius of 767.81 feet. What is the altitude of the bisector of the chord to the circle? +2 if you can tell me how you come up with the answer.
Just Jeepin'
> Rusty Vandura - www.tinyurl.com/keepoppo
05/23/2019 at 20:38 |
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I vote 16’, since the two extremes of the function are the points where x=r and thus the length of the chord equals r, and x=0 when the length of the chord equals 0.
I’m sure there’s calculus to prove me wrong, but FAKE NEWS.
ttyymmnn
> Rusty Vandura - www.tinyurl.com/keepoppo
05/23/2019 at 20:40 |
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someassemblyrequired
> Rusty Vandura - www.tinyurl.com/keepoppo
05/23/2019 at 20:41 |
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https://www.mathopenref.com/sagitta.html
- it explains it and provides a calculator - the bisector in your problem is the “sagitta”
EngineerWithTools
> Rusty Vandura - www.tinyurl.com/keepoppo
05/23/2019 at 20:42 |
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.04 ft
ttyymmnn
> Rusty Vandura - www.tinyurl.com/keepoppo
05/23/2019 at 20:42 |
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Just Jeepin'
> someassemblyrequired
05/23/2019 at 20:43 |
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As a Sagittarius I’m more than a little disappointed I didn’t know that term.
someassemblyrequired
> Just Jeepin'
05/23/2019 at 20:44 |
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that’s the MathNerd ™ term, it’s pretty obscure
Levitas
> Rusty Vandura - www.tinyurl.com/keepoppo
05/23/2019 at 20:44 |
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If the chord connects on either end to the circle, then you have a right triangle with hypotenuse with the same length as your radius. Using half the length of the chord (8') and R (767.81'), you can solve for the third leg using Pythagoreum’s theorem, which would be equal to (R - x).
So then (R)^2 = (8)^2 + (R-x)^2, solve for x.
x = .0417'
ed: damn too slow
TorqueToYield
> Rusty Vandura - www.tinyurl.com/keepoppo
05/23/2019 at 20:46 |
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hmm
Future Heap Owner
> Just Jeepin'
05/23/2019 at 20:49 |
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Not all paths between two points are a straight line. In fact, to a very good approximation, none of them are. REAL NUMBERS
OPPOsaurus WRX
> Rusty Vandura - www.tinyurl.com/keepoppo
05/23/2019 at 20:55 |
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Id just draw it in CAD and measure it but that would mean id have to get up
MrDakka
> OPPOsaurus WRX
05/23/2019 at 21:07 |
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Gotta love constraints and the measure tool
Rusty Vandura - www.tinyurl.com/keepoppo
> Levitas
05/23/2019 at 23:06 |
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Yes! Indeed.
Rusty Vandura - www.tinyurl.com/keepoppo
> EngineerWithTools
05/23/2019 at 23:09 |
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I did that in Algebra once. Or was it Calculus? The math is familiar to me, but it's been many years. Thank you!
415s30 W123TSXWaggoIIIIIIo ( •_•))°)
> Rusty Vandura - www.tinyurl.com/keepoppo
05/23/2019 at 23:31 |
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Distraxi's idea of perfection is a Jagroen
> Future Heap Owner
05/23/2019 at 23:46 |
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!!! UNKNOWN CONTENT TYPE !!!
Maybe no t in your hyperspace.
Rusty Vandura - www.tinyurl.com/keepoppo
> 415s30 W123TSXWaggoIIIIIIo ( •_•))°)
05/24/2019 at 00:37 |
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Highlander-Datsuns are Forever
> EngineerWithTools
05/24/2019 at 01:07 |
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It takes an engineer! They can’t get ride of us so easily.
ateamfan42
> EngineerWithTools
05/24/2019 at 09:44 |
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Pythagoras strikes again!
ttyymmnn
> Rusty Vandura - www.tinyurl.com/keepoppo
05/24/2019 at 11:33 |
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Was that a practical problem or an intellectual problem?
vicali
> 415s30 W123TSXWaggoIIIIIIo ( •_•))°)
05/24/2019 at 12:10 |
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Rusty Vandura - www.tinyurl.com/keepoppo
> ttyymmnn
05/24/2019 at 12:26 |
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Practical. My BIL is building something for his job, probably a bridge, and he’ll be looking for how high the crest is from either side. Probably ordering some ready-made fabricated steel structure. But Oppo came through and showed me the geometry, and it’s really quite accessible.
AMGtech - now with more recalls!
> Rusty Vandura - www.tinyurl.com/keepoppo
05/24/2019 at 16:14 |
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I’m just trying to figure out how you measure music in feet
Rusty Vandura - www.tinyurl.com/keepoppo
> AMGtech - now with more recalls!
05/24/2019 at 16:18 |
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Yes, an interesting notion how that word shows up in both contexts. I’ll ask Ttyymmnn about that one...