4 Revolutionary Riddles Resolved!

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+4 1. Burimi commented 31 days ago

Thanks Geek for publishing this video! I think Thundersnow will be very happy!
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+4 2. thundersnow commented 31 days ago

Thundersnow is very happy! :) <3 <3 <3
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+2 3. ComentAtor commented 31 days ago

why would you put not one but two ping pong balls in a perfecty good jar of honey?? :D :squirrel:
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+1 4. thundersnow commented 31 days ago

#3 Where one ball goes, the other one wants to be. O:)
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+2 5. ComentAtor commented 31 days ago

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+1 6. thundersnow commented 31 days ago

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+2 7. MindTrick commented 31 days ago

I got every one right! Im just kidding... the race track issue seemed so clear to me first, even tho i knew there was more to it, it surprised me a bit. The viscosity i sort of had correct, just not the exact content... cool riddles tho, a bit higher level than "the person on the street" will manage to solve, but hey, that's how we learn, aint it? :)
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+1 8. thundersnow commented 31 days ago

#7 You got it! To be honest, I don't even totally get the bike and race track concepts yet, just partially.
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+1 9. kirkelicious commented 30 days ago

#8 It gets easier to understand if you imagine the pedals to be another wheel that is fixed on the same axis as the back wheel of the bike. Forget about the chain and the gears for a moment. You could just as well change the diameter of the "pedal wheel"to alter the transmission ratio. Change the lever length - change the torque. Simple mechanics.
In a very low gear the Pedal wheel would be bigger than the real wheel, which conveniently leaves with the model of the train problem. You pull the pedal backwards, the bike rolls forwards.
If this pedal wheel has a smaller diameter the direction of motion is the same for the pedal and the wheel. therefore pulling it backwards makes the whole system slide backwards.

The Track Riddle can solved analytically:

With the track length s and the time t the average speed for the first lap is

twice that speed would be
2*s / t1

The average time for two laps is
(s+s) / (t1+t2) = 2*s / (t1+t2)

you can write that as
2*v1 = 2*s / t1 =2*s / (t1+t2)

that means

solve that for t2:
t2 = t1-t1 =0

You have no time left to complete lap nr.2
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+2 10. thundersnow commented 30 days ago

#9 I really appreciate your explanation, and I'm trying to understand it, but I have to confess, that you are dealing with someone who doesn't even know the basic mechanics of a bicycle. I just realized that myself when I started reading your comment, and I understand English very well, but I still couldn't understand the content of your comment (and I'm pretty sure I'm not currently stroking and presenting with receptive aphasia)....so I realized in order to understand your explanation, which I want to understand, I need to first understand the basic mechanics of a bicycle, especially those parts down there by the pedal, wherever the pedal is attached to, so I started googling it, it's actually very interesting, and I like learning something new like that. Thank you, kirkelicious for your patience. <3.....and I haven't even started to understand the other explanations of yours....so lemme make a wild guess, you are a physics professor at a university somewhere here or in Europe or at least a physics teacher at a high school (or Gymnasium) :)....

These are my sources:
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+2 11. kirkelicious commented 30 days ago

#10 not quite a Professor, no :)
I did work in the Physics-Department of an University in Europe, though. Nice guess.
Currently I am developing industrial CT-scanners as a computational engineer. Got to know your physics for that...

Don't worry to underststand this substitute wheel stuff I dreamt up, if mechanics is unknown territory for you. Not sure if my explanation even made it harder to understand ;) Some graphics would help a lot.

My explanation of the track riddle is basic 8th grade maths. Don't be scared of some algebra. Come on Thunders, you can figure it out :)
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+1 12. thundersnow commented 30 days ago

#11 It's okay, like I mentioned I like to learn new things and I also like the challenge, makes me stubborn to keep pursuing it. And yes, anything visual helps me greatly, and actually having it in front of me and someone explaining it, basically spoon feeding me ;), works the best.
The track is 8th grade math?!! Oh no, how far behind have I remained! :'( Your cheering on helps, but the pressure is now mounting....
"..developing industrial CT scanners..." Wow, just wow! Now there is something a little more complicated than a bike pedal. The most I know of CT scanners is that patients get CTs done with such a frequency, sometimes unnecessary follow ups, which is kind of a concern, considering how much X-rays they're exposed to, so much more than a conventional X-ray, but amazing diagnostics they are. I get to read the results. Also love how they sound like a spinning washing machine while they are running....okay, time to get back to my bike pedals :)
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+1 13. sp176 commented 30 days ago

The flaw in the race track logic is how he is trying to calculate speed. He uses 2*V = 2*D/T. Well, the race track length isn't changing. The distance is not doubling. The proper form of the equation should be 2*V=D/(0.5*T). You run the same distance, but in half the time. It is very easy to calculate that V2 = 3*V1, to get your average speed over both laps. Now, if you try to say that one of the speeds is weighted more because of less time over the second lap, well then you need to do some integral calculus, which is also not difficult. Just make sure you use realize that it is time per lap changing, not the distance.
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+1 14. kirkelicious commented 29 days ago

#13 You are over-complicating things.

Fist of all: 2*V=D/(0.5*T) and 2*V = 2*D/T is exactly the same thing (1/0,5=2) ;)

The question asks for the average speed for one or two laps repectively, with each lap run with a constant speed as calculated by (total distance)/(total time).

It is really unambiguously worded.

after one lap that average speed is D/T1 and after the second it's 2D/(T1+T2)

That leaves you with the problem

2D/T1 = 2D/(T1+T2)

solve that for T2 and you realize that you're screwed

Lets follow your logic:
V2 = 3*V1


that means

And the average speed becomes
Va= 2D/(T1+T1/3)= 2D/(4/3*T1) = 3/2*(D/T1)

but we are looking for Va=2V1=2D/T1