>A broken clock is right two times every day

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>A broken clock is right two times every day

What if my clock (B) was broken in such a way that it ran at 1.5x the speed of a reference clock (A)? How many times on average would it he correct every day?

What if my clock (B) ran in reverse at half the speed of clock (A)?

What if clock (B) accelerated over time at a rate of X^2?

Can you generalize this into a formula to determine the maximum, minimum, and average number of times clock (B) would be correct with respect to clock (A) on any given day?

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Broken clock usually means clock that doesn't run at all though

The rest of your question is just high school level physics problem, and probably your homework

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>>7755688

Is that your way of saying you can't solve the problems?

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The difference of amounts of revolutions per day is equal to the number of intersection as long as they have constant speed.

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>>7755623

just simulate it with mcmc in matlab

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>>7755693

It's my way of saying I won't solve your homework problems

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>>7755702

Op here, I haven't had homework for over 4 years. This was just a thought experiment that popped into my head after someone used that saying in conversation with me earlier today.

I've only got as far as finding solutions for the set of equations, e.g. For half speed

A=2B

A%12=B%12

0<=A<=24

0<=B

There are two solutions, A=0,24 B=0,12

So the clock is right twice a day. It gets more difficult when you try to make a meta formula that can count the number of solutions. It's even more challenging when you introduce acceleration on clock B as you can't restrict A/B to 24 hours because they get out of phase

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>>7755714

its not really that hard , just integrate your time formula minus your constant speed rotation formula to find the number of rotations

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>>7755736

\operatorname{mod}\left(x,12\right)\ \left\{0\le x\right\}

\operatorname{mod}\left(f\left(x\right),12\right)\ \left\{0\le x\right\}

\operatorname{abs}\left(\operatorname{floor}\left(\frac{x-f\left(x\right)}{12}\right)\right)\left\{0<x\right\}

case 1

f\left(x\right)=1.5x

case 2

f\left(x\right)=-0.5x

case 3

f\left(x\right)=x^2

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>>7755827

oops, I dont know how to format equations, just paste the lines into https://www.desmos.com/calculator

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>>7755829

that sure worked like a charm

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>>7755840

You need to paste each line one by one

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>>7755840

Or I can just save it

https://www.desmos.com/calculator/iwzya4jk4y

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>>7755841

ok better, ty

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>>7755850

I dont understand... It works perfectly in my browser...

Try the link

https://www.desmos.com/calculator/iwzya4jk4y

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>>7755850

Oh I do understand, you cant have f(x)=1.5x and f(x)=-0.5x at the same time, delete on of them

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>>7755827

yeah something like that

you probably want to convert into circular coordinates and just use e^i \theta

if you allow the function to go backwards and forwards you probably need to use an absolute value metric and some kind of indicator function for counting

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>>7755623

yes

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>>7756093

/thread

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>>7755623

That's not what a broken clock means you dipshit

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>>7759485

dictionary.com

[broh-kuh n]

verb

4. not functioning properly; out of working order.

>functioning, but not properly

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