Bike Myths We Wish Would Die
02-22-23, 01:23 PM
#551
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Why don't you put some of your physics knowledge on display and calculate some numbers for us? You can start with something simple: pick two wheel sets and calculate how many extra watts are required to bring the heavier set up to speed when accelerating from a full stop.
I = (1/2) * m * r^2
where I is the rotational inertia, m is the mass of the cylinder, and r is the radius of the cylinder.
Before the change, the mass of the wheel can be calculated using its density and volume:
m = density * volume
The density of the wheel can be assumed to be 2700 kg/m^3, which is the density of aluminum alloy commonly used for bicycle wheels. The volume of the wheel can be calculated using its diameter:
V = (1/4) * pi * d^2 * t
where V is the volume, d is the diameter, and t is the thickness of the wheel.
Let's assume the wheel has a thickness of 25mm. Substituting the values, we get:
V = (1/4) * pi * (0.622m)^2 * 0.025m = 0.0048 m^3
m = 2700 kg/m^3 * 0.0048 m^3 = 12.96 kg
The radius of the wheel is half of its diameter:
r = d/2 = 0.311m
Using the formula for rotational inertia, we can calculate the rotational inertia of the wheel before the change:
I_before = (1/2) * 12.96 kg * (0.311m)^2 = 0.619 kg*m^2
Now, let's add or subtract 453g (0.453kg) from the rims of the wheel. This will change the mass of the wheel to:
m_after = 12.96 kg + 0.453 kg = 13.413 kg
Using the same formula, we can calculate the rotational inertia of the wheel after the change:
I_after = (1/2) * 13.413 kg * (0.311m)^2 = 0.641 kg*m^2
The difference in the rotational inertia of the wheel before and after the change is:
ΔI = I_after - I_before = 0.641 kgm^2 - 0.619 kgm^2 = 0.022 kg*m^2
So, adding or subtracting 453g from the rims of the wheel will result in a difference of 0.022 kg*m^2 in the wheel's rotational inertia.
Whether a change in the rotational inertia of a wheel is significant or not depends on the application and the level of precision required. In the case of a bicycle, a change of 0.022 kg*m^2 in the wheel's rotational inertia may not have a noticeable effect on the overall performance of the bike.
For professional cyclists or competitive athletes where small differences in performance can make a significant impact, a change in the rotational inertia of the wheels could very well have a noticeable effect on the bike's acceleration, speed, and handling.
Thus, we have shown that reducing the rotational inertia of the wheels can improve the bike's performance by making it easier to accelerate and change direction. The degree to which this is true is dependent on the amount of change, the characteristics of the ride, and the individual rider's sensitivity to differences in bike handling.
Last edited by Jeff Neese; 02-22-23 at 01:40 PM.
02-22-23, 01:33 PM
#552
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02-22-23, 01:41 PM
#553
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wrong and has been posted many time here is company that specializes in industrial wax lubricants..... especially for industrial chains
https://www.klueber.com/us/en/produc...icating-waxes/
https://www.klueber.com/us/en/produc...icating-waxes/
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02-22-23, 02:00 PM
#554
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There is no inertia at a constant 40 mph.
I think you mean 406 wheel size BTW.
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02-22-23, 02:03 PM
#555
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Why low 13's vs high 12's. Same compound. Just much wider. Same height.
Nobody ever keeps tire size the same when seeking more traction.
Edit: better yet, why does everyone go to bigger brake pads to increase motorcar stopping force. (Oh yes, we need infinitely smooth surfaces besides a vacuum)
Last edited by GhostRider62; 02-22-23 at 02:07 PM.
02-22-23, 02:13 PM
#556
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Let's assume we have a 622mm diameter wheel (700c) and we add or subtract 453g (1 pound) from the rims of the wheel. We can use the formula for the rotational inertia of a solid cylinder to calculate the difference in the rotational inertia of the wheel before and after the change:
I = (1/2) * m * r^2
where I is the rotational inertia, m is the mass of the cylinder, and r is the radius of the cylinder.
Before the change, the mass of the wheel can be calculated using its density and volume:
m = density * volume
The density of the wheel can be assumed to be 2700 kg/m^3, which is the density of aluminum alloy commonly used for bicycle wheels. The volume of the wheel can be calculated using its diameter:
V = (1/4) * pi * d^2 * t
where V is the volume, d is the diameter, and t is the thickness of the wheel.
Let's assume the wheel has a thickness of 25mm. Substituting the values, we get:
V = (1/4) * pi * (0.622m)^2 * 0.025m = 0.0048 m^3
m = 2700 kg/m^3 * 0.0048 m^3 = 12.96 kg
The radius of the wheel is half of its diameter:
r = d/2 = 0.311m
Using the formula for rotational inertia, we can calculate the rotational inertia of the wheel before the change:
I_before = (1/2) * 12.96 kg * (0.311m)^2 = 0.619 kg*m^2
Now, let's add or subtract 453g (0.453kg) from the rims of the wheel. This will change the mass of the wheel to:
m_after = 12.96 kg + 0.453 kg = 13.413 kg
Using the same formula, we can calculate the rotational inertia of the wheel after the change:
I_after = (1/2) * 13.413 kg * (0.311m)^2 = 0.641 kg*m^2
The difference in the rotational inertia of the wheel before and after the change is:
ΔI = I_after - I_before = 0.641 kgm^2 - 0.619 kgm^2 = 0.022 kg*m^2
So, adding or subtracting 453g from the rims of the wheel will result in a difference of 0.022 kg*m^2 in the wheel's rotational inertia.
Whether a change in the rotational inertia of a wheel is significant or not depends on the application and the level of precision required. In the case of a bicycle, a change of 0.022 kg*m^2 in the wheel's rotational inertia may not have a noticeable effect on the overall performance of the bike.
For professional cyclists or competitive athletes where small differences in performance can make a significant impact, a change in the rotational inertia of the wheels could very well have a noticeable effect on the bike's acceleration, speed, and handling.
Thus, we have shown that reducing the rotational inertia of the wheels can improve the bike's performance by making it easier to accelerate and change direction. The degree to which this is true is dependent on the amount of change, the characteristics of the ride, and the individual rider's sensitivity to differences in bike handling.
I = (1/2) * m * r^2
where I is the rotational inertia, m is the mass of the cylinder, and r is the radius of the cylinder.
Before the change, the mass of the wheel can be calculated using its density and volume:
m = density * volume
The density of the wheel can be assumed to be 2700 kg/m^3, which is the density of aluminum alloy commonly used for bicycle wheels. The volume of the wheel can be calculated using its diameter:
V = (1/4) * pi * d^2 * t
where V is the volume, d is the diameter, and t is the thickness of the wheel.
Let's assume the wheel has a thickness of 25mm. Substituting the values, we get:
V = (1/4) * pi * (0.622m)^2 * 0.025m = 0.0048 m^3
m = 2700 kg/m^3 * 0.0048 m^3 = 12.96 kg
The radius of the wheel is half of its diameter:
r = d/2 = 0.311m
Using the formula for rotational inertia, we can calculate the rotational inertia of the wheel before the change:
I_before = (1/2) * 12.96 kg * (0.311m)^2 = 0.619 kg*m^2
Now, let's add or subtract 453g (0.453kg) from the rims of the wheel. This will change the mass of the wheel to:
m_after = 12.96 kg + 0.453 kg = 13.413 kg
Using the same formula, we can calculate the rotational inertia of the wheel after the change:
I_after = (1/2) * 13.413 kg * (0.311m)^2 = 0.641 kg*m^2
The difference in the rotational inertia of the wheel before and after the change is:
ΔI = I_after - I_before = 0.641 kgm^2 - 0.619 kgm^2 = 0.022 kg*m^2
So, adding or subtracting 453g from the rims of the wheel will result in a difference of 0.022 kg*m^2 in the wheel's rotational inertia.
Whether a change in the rotational inertia of a wheel is significant or not depends on the application and the level of precision required. In the case of a bicycle, a change of 0.022 kg*m^2 in the wheel's rotational inertia may not have a noticeable effect on the overall performance of the bike.
For professional cyclists or competitive athletes where small differences in performance can make a significant impact, a change in the rotational inertia of the wheels could very well have a noticeable effect on the bike's acceleration, speed, and handling.
Thus, we have shown that reducing the rotational inertia of the wheels can improve the bike's performance by making it easier to accelerate and change direction. The degree to which this is true is dependent on the amount of change, the characteristics of the ride, and the individual rider's sensitivity to differences in bike handling.
Sad.
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02-22-23, 02:16 PM
#557
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Countersteering is actually the only way you can lean into a corner. Explained here with a video:-
https://www.renehersecycles.com/myth...untersteering/
We all counter steer instinctively and it would be impossible to lean into a turn if we didn't. With a very heavy bike (like a motorbike) we might have to consciously counter steer. Which is probably why it gets discussed more by bikers.
https://www.renehersecycles.com/myth...untersteering/
We all counter steer instinctively and it would be impossible to lean into a turn if we didn't. With a very heavy bike (like a motorbike) we might have to consciously counter steer. Which is probably why it gets discussed more by bikers.
I respectfully disagree. When you turn a bike while riding no hands, countersteering cannot be the initial event, as you have no direct input to the steering mechanism (handlebars)
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02-22-23, 02:21 PM
#558
Full Member
So splain me this.
Why low 13's vs high 12's. Same compound. Just much wider. Same height.
Nobody ever keeps tire size the same when seeking more traction.
Edit: better yet, why does everyone go to bigger brake pads to increase motorcar stopping force. (Oh yes, we need infinitely smooth surfaces besides a vacuum)
Why low 13's vs high 12's. Same compound. Just much wider. Same height.
Nobody ever keeps tire size the same when seeking more traction.
Edit: better yet, why does everyone go to bigger brake pads to increase motorcar stopping force. (Oh yes, we need infinitely smooth surfaces besides a vacuum)
02-22-23, 02:29 PM
#559
Senior Member
Let's assume we have a 622mm diameter wheel (700c) and we add or subtract 453g (1 pound) from the rims of the wheel. We can use the formula for the rotational inertia of a solid cylinder to calculate the difference in the rotational inertia of the wheel before and after the change:
I = (1/2) * m * r^2
where I is the rotational inertia, m is the mass of the cylinder, and r is the radius of the cylinder.
Before the change, the mass of the wheel can be calculated using its density and volume:
m = density * volume
The density of the wheel can be assumed to be 2700 kg/m^3, which is the density of aluminum alloy commonly used for bicycle wheels. The volume of the wheel can be calculated using its diameter:
V = (1/4) * pi * d^2 * t
where V is the volume, d is the diameter, and t is the thickness of the wheel.
Let's assume the wheel has a thickness of 25mm. Substituting the values, we get:
V = (1/4) * pi * (0.622m)^2 * 0.025m = 0.0048 m^3
m = 2700 kg/m^3 * 0.0048 m^3 = 12.96 kg
The radius of the wheel is half of its diameter:
r = d/2 = 0.311m
Using the formula for rotational inertia, we can calculate the rotational inertia of the wheel before the change:
I_before = (1/2) * 12.96 kg * (0.311m)^2 = 0.619 kg*m^2
Now, let's add or subtract 453g (0.453kg) from the rims of the wheel. This will change the mass of the wheel to:
m_after = 12.96 kg + 0.453 kg = 13.413 kg
Using the same formula, we can calculate the rotational inertia of the wheel after the change:
I_after = (1/2) * 13.413 kg * (0.311m)^2 = 0.641 kg*m^2
The difference in the rotational inertia of the wheel before and after the change is:
ΔI = I_after - I_before = 0.641 kgm^2 - 0.619 kgm^2 = 0.022 kg*m^2
So, adding or subtracting 453g from the rims of the wheel will result in a difference of 0.022 kg*m^2 in the wheel's rotational inertia.
Whether a change in the rotational inertia of a wheel is significant or not depends on the application and the level of precision required. In the case of a bicycle, a change of 0.022 kg*m^2 in the wheel's rotational inertia may not have a noticeable effect on the overall performance of the bike.
For professional cyclists or competitive athletes where small differences in performance can make a significant impact, a change in the rotational inertia of the wheels could very well have a noticeable effect on the bike's acceleration, speed, and handling.
Thus, we have shown that reducing the rotational inertia of the wheels can improve the bike's performance by making it easier to accelerate and change direction. The degree to which this is true is dependent on the amount of change, the characteristics of the ride, and the individual rider's sensitivity to differences in bike handling.
I = (1/2) * m * r^2
where I is the rotational inertia, m is the mass of the cylinder, and r is the radius of the cylinder.
Before the change, the mass of the wheel can be calculated using its density and volume:
m = density * volume
The density of the wheel can be assumed to be 2700 kg/m^3, which is the density of aluminum alloy commonly used for bicycle wheels. The volume of the wheel can be calculated using its diameter:
V = (1/4) * pi * d^2 * t
where V is the volume, d is the diameter, and t is the thickness of the wheel.
Let's assume the wheel has a thickness of 25mm. Substituting the values, we get:
V = (1/4) * pi * (0.622m)^2 * 0.025m = 0.0048 m^3
m = 2700 kg/m^3 * 0.0048 m^3 = 12.96 kg
The radius of the wheel is half of its diameter:
r = d/2 = 0.311m
Using the formula for rotational inertia, we can calculate the rotational inertia of the wheel before the change:
I_before = (1/2) * 12.96 kg * (0.311m)^2 = 0.619 kg*m^2
Now, let's add or subtract 453g (0.453kg) from the rims of the wheel. This will change the mass of the wheel to:
m_after = 12.96 kg + 0.453 kg = 13.413 kg
Using the same formula, we can calculate the rotational inertia of the wheel after the change:
I_after = (1/2) * 13.413 kg * (0.311m)^2 = 0.641 kg*m^2
The difference in the rotational inertia of the wheel before and after the change is:
ΔI = I_after - I_before = 0.641 kgm^2 - 0.619 kgm^2 = 0.022 kg*m^2
So, adding or subtracting 453g from the rims of the wheel will result in a difference of 0.022 kg*m^2 in the wheel's rotational inertia.
Whether a change in the rotational inertia of a wheel is significant or not depends on the application and the level of precision required. In the case of a bicycle, a change of 0.022 kg*m^2 in the wheel's rotational inertia may not have a noticeable effect on the overall performance of the bike.
For professional cyclists or competitive athletes where small differences in performance can make a significant impact, a change in the rotational inertia of the wheels could very well have a noticeable effect on the bike's acceleration, speed, and handling.
Thus, we have shown that reducing the rotational inertia of the wheels can improve the bike's performance by making it easier to accelerate and change direction. The degree to which this is true is dependent on the amount of change, the characteristics of the ride, and the individual rider's sensitivity to differences in bike handling.
Why are you modeling the wheel as a whole at all if your end calculation is just about the difference from a change in rim mass?
Why do your calculations pertaining to mass added "at the rim" distribute it evenly across the entire disk?
Why is this written in the style of a ChatGPT essay?
02-22-23, 02:43 PM
#560
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Why are you modeling a wheel as a huge 29-pound aluminum disk?
Why are you modeling the wheel as a whole at all if your end calculation is just about the difference from a change in rim mass?
Why do your calculations pertaining to mass added "at the rim" distribute it evenly across the entire disk?
Why is this written in the style of a ChatGPT essay?
Why are you modeling the wheel as a whole at all if your end calculation is just about the difference from a change in rim mass?
Why do your calculations pertaining to mass added "at the rim" distribute it evenly across the entire disk?
Why is this written in the style of a ChatGPT essay?
02-22-23, 02:50 PM
#561
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There are 3 dead people, who ignored my advices.(age 72, 60, 47) I was trying to save them from heart attack and stroke, they didn't listen. 4th one had heart attack but didn't die. They probably didn't listen because I was not a doctor. RIP, Alex, James and Dirk!!! I tried!!!
Yeah, that happened. Sure.
Are you a duck? I hear a quack.
02-22-23, 02:55 PM
#562
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Looking forward to the AI-generated response from the guy who got ChatGPT to do his homework.
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02-22-23, 02:59 PM
#563
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A valiant effort. Unfortunately, a wall of text/formulae that is 91.62% AI Bot-generated* really doesn't provide evidence of anything at all, especially given that the 'essay' you've had an AI program write for you doesn't address the question tomato coupe posed and to which you were ostensibly responding.
Sad.
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Sad.
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Edit- He just got back to me. He says zerogpt only catches the real amateur/common efforts. They are now implementing steps to make AI much less of a concern in grading (didn't elaborate, looking forward to the discussion!)), and on campus they are seeing AI that zerogpt can't pick up on. He says that you cannot rely solely on an AI detector. Anyway, carry on.
Last edited by Erzulis Boat; 02-22-23 at 03:12 PM.
02-22-23, 03:00 PM
#564
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Why are you modeling a wheel as a huge 29-pound aluminum disk?
Why are you modeling the wheel as a whole at all if your end calculation is just about the difference from a change in rim mass?
Why do your calculations pertaining to mass added "at the rim" distribute it evenly across the entire disk?
Why is this written in the style of a ChatGPT essay?
Why are you modeling the wheel as a whole at all if your end calculation is just about the difference from a change in rim mass?
Why do your calculations pertaining to mass added "at the rim" distribute it evenly across the entire disk?
Why is this written in the style of a ChatGPT essay?
I almost feel sorry for the guy. I'm a total math idiot, and I wouldn't know if I was posting a wall of nonsense. I sure wouldn't try to fool anyone else with it.
02-22-23, 03:05 PM
#565
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Academic integrity is a core value of education, and it is crucial for personal growth and professional success. When you submit work that is not your own, you undermine the trust that others have in you and risk damaging your reputation and credibility. In addition, cheating in any form can result in severe consequences, such as failing the assignment or the course, being expelled from the school or university, or even facing legal action.
Furthermore, it is important to recognize the value of learning and growing from your mistakes. If you feel that you are struggling with a particular topic or assignment, it is better to seek help from your teacher, tutor, or classmates. There are many resources available to students to help them improve their skills and knowledge, such as writing centers, academic advisors, and online tutorials.
In conclusion, it is never justifiable to present someone else's work as your own, regardless of the circumstances. Academic integrity is an essential aspect of education, and it is crucial to maintain it at all times. If you feel that you are struggling with an assignment or topic, seek help and guidance from appropriate sources, rather than resorting to unethical practices.
02-22-23, 03:07 PM
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My university has yet to develop a 'policy' on the issues raised, but my fear is that it won't be pretty. We already have 'student lawyers' and distinguished university faculty members lobbying for the 'embrace the tech' approach: aid to thought/'creativity' etc. blah blah blah -- 'we shall all have an AI Bot on our shoulders.'
Can't wait to retire.
Last edited by badger1; 02-22-23 at 03:15 PM.
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02-22-23, 03:09 PM
#567
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I'm sorry, but as an AI language model, I cannot help you in writing an essay that justifies academic dishonesty or plagiarism. It is not ethical or fair to present someone else's work as your own, regardless of whether it was generated by an AI or not.
Academic integrity is a core value of education, and it is crucial for personal growth and professional success. When you submit work that is not your own, you undermine the trust that others have in you and risk damaging your reputation and credibility. In addition, cheating in any form can result in severe consequences, such as failing the assignment or the course, being expelled from the school or university, or even facing legal action.
Furthermore, it is important to recognize the value of learning and growing from your mistakes. If you feel that you are struggling with a particular topic or assignment, it is better to seek help from your teacher, tutor, or classmates. There are many resources available to students to help them improve their skills and knowledge, such as writing centers, academic advisors, and online tutorials.
In conclusion, it is never justifiable to present someone else's work as your own, regardless of the circumstances. Academic integrity is an essential aspect of education, and it is crucial to maintain it at all times. If you feel that you are struggling with an assignment or topic, seek help and guidance from appropriate sources, rather than resorting to unethical practices.
Academic integrity is a core value of education, and it is crucial for personal growth and professional success. When you submit work that is not your own, you undermine the trust that others have in you and risk damaging your reputation and credibility. In addition, cheating in any form can result in severe consequences, such as failing the assignment or the course, being expelled from the school or university, or even facing legal action.
Furthermore, it is important to recognize the value of learning and growing from your mistakes. If you feel that you are struggling with a particular topic or assignment, it is better to seek help from your teacher, tutor, or classmates. There are many resources available to students to help them improve their skills and knowledge, such as writing centers, academic advisors, and online tutorials.
In conclusion, it is never justifiable to present someone else's work as your own, regardless of the circumstances. Academic integrity is an essential aspect of education, and it is crucial to maintain it at all times. If you feel that you are struggling with an assignment or topic, seek help and guidance from appropriate sources, rather than resorting to unethical practices.
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02-22-23, 03:16 PM
#568
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A valiant effort. Unfortunately, a wall of text/formulae that is 91.62% AI Bot-generated* really doesn't provide evidence of anything at all, especially given that the 'essay' you've had an AI program write for you doesn't address the question tomato coupe posed and to which you were ostensibly responding.
Sad.
*22/02/2023 AI Text Detector (Zero GPT: www.zerogpt.com)
Sad.
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If we want to accelerated from 0 to 120 rpm in 10 seconds, the difference in power required to accelerate the two wheels from 0 to 120 rpm is approximately 1.02 Watts.
For the wheel with a change of 453g:ΔK = 48.8 Joules (calculated earlier) t = 10 seconds
P = ΔK / t = 48.8 J / 10 s = 4.88 Watts
For the wheel with a change of 906g:
ΔK = 59.0 Joules (calculated earlier) t = 10 seconds
P = ΔK / t = 59.0 J / 10 s = 5.90 Watts
And what do you mean "a valiant effort"? He was a friggin' genius, especially considering it was the 17th century. He might have even been smarter than the guy from Cambridge with the YouTube video.
Last edited by Jeff Neese; 02-22-23 at 03:23 PM.
02-22-23, 03:23 PM
#569
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Oh, you mean wattage required for aceleration? OK, I'll bite. I'll use a random example of going from full stop to 120rpm in 10 seconds.
If we want to accelerated from 0 to 120 rpm in 10 seconds, the difference in power required to accelerate the two wheels from 0 to 120 rpm is approximately 1.02 Watts.
For the wheel with a change of 453g:ΔK = 48.8 Joules (calculated earlier) t = 10 seconds
P = ΔK / t = 48.8 J / 10 s = 4.88 Watts
For the wheel with a change of 906g:
ΔK = 59.0 Joules (calculated earlier) t = 10 seconds
P = ΔK / t = 59.0 J / 10 s = 5.90 Watts
And what do you mean "a valiant effort"? He was a friggin' genius, especially considering it was the 17th century.
If we want to accelerated from 0 to 120 rpm in 10 seconds, the difference in power required to accelerate the two wheels from 0 to 120 rpm is approximately 1.02 Watts.
For the wheel with a change of 453g:ΔK = 48.8 Joules (calculated earlier) t = 10 seconds
P = ΔK / t = 48.8 J / 10 s = 4.88 Watts
For the wheel with a change of 906g:
ΔK = 59.0 Joules (calculated earlier) t = 10 seconds
P = ΔK / t = 59.0 J / 10 s = 5.90 Watts
And what do you mean "a valiant effort"? He was a friggin' genius, especially considering it was the 17th century.
My earlier comment was simply meant to highlight your valiant, but intellectually dishonest, effort. Your post that I submitted to analysis was composed mainly by an AI bot program, not you. That is intellectual dishonesty, in my view, and was the only thing I found interesting about that post.
Bye.
02-22-23, 04:12 PM
#570
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Oh, you mean wattage required for aceleration? OK, I'll bite. I'll use a random example of going from full stop to 120rpm in 10 seconds.
If we want to accelerated from 0 to 120 rpm in 10 seconds, the difference in power required to accelerate the two wheels from 0 to 120 rpm is approximately 1.02 Watts.
For the wheel with a change of 453g:ΔK = 48.8 Joules (calculated earlier) t = 10 seconds
P = ΔK / t = 48.8 J / 10 s = 4.88 Watts
For the wheel with a change of 906g:
ΔK = 59.0 Joules (calculated earlier) t = 10 seconds
P = ΔK / t = 59.0 J / 10 s = 5.90 Watts
And what do you mean "a valiant effort"? He was a friggin' genius, especially considering it was the 17th century. He might have even been smarter than the guy from Cambridge with the YouTube video.
If we want to accelerated from 0 to 120 rpm in 10 seconds, the difference in power required to accelerate the two wheels from 0 to 120 rpm is approximately 1.02 Watts.
For the wheel with a change of 453g:ΔK = 48.8 Joules (calculated earlier) t = 10 seconds
P = ΔK / t = 48.8 J / 10 s = 4.88 Watts
For the wheel with a change of 906g:
ΔK = 59.0 Joules (calculated earlier) t = 10 seconds
P = ΔK / t = 59.0 J / 10 s = 5.90 Watts
And what do you mean "a valiant effort"? He was a friggin' genius, especially considering it was the 17th century. He might have even been smarter than the guy from Cambridge with the YouTube video.
02-22-23, 04:28 PM
#571
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Your numbers and wiggles and squibbles mean absolutely nothing to me. Address them to those who understand these things.
My earlier comment was simply meant to highlight your valiant, but intellectually dishonest, effort. Your post that I submitted to analysis was composed mainly by an AI bot program, not you. That is intellectual dishonesty, in my view, and was the only thing I found interesting about that post.
Bye.
My earlier comment was simply meant to highlight your valiant, but intellectually dishonest, effort. Your post that I submitted to analysis was composed mainly by an AI bot program, not you. That is intellectual dishonesty, in my view, and was the only thing I found interesting about that post.
Bye.
02-22-23, 04:32 PM
#572
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There are a lot of ways to look things up. Frankly, the person that made the challenge could just as easily have done the same. It's in textbooks and would be found through Google. He asked for the calculations, and I gave them to him. I guess he was just being lazy and wanted me to look them up for him. Or maybe he tried but couldn't find it. Doesn't matter - there's the math.
02-22-23, 04:42 PM
#573
Senior Member
No you didn't. The answers you gave are fairly meaningless, since they include neither your assumptions nor most of the relevant calculations.
As it happens, the specific answer you offered (10.2J over ten seconds) is correct if the added mass is placed at a diameter of 755mm. Which is kind of a strange assumption, as is your use of rotational velocity as an alias for speed at the opening of the problem.
Yes. And there are a lot of ways to fail to comprehend or meaningfully use the material that you look up.
As it happens, the specific answer you offered (10.2J over ten seconds) is correct if the added mass is placed at a diameter of 755mm. Which is kind of a strange assumption, as is your use of rotational velocity as an alias for speed at the opening of the problem.
There are a lot of ways to look things up.
Yes. And there are a lot of ways to fail to comprehend or meaningfully use the material that you look up.
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02-22-23, 04:45 PM
#574
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What is the definition of "dangerous"? Does something need to be totally free of risk to be "safe"? How is cycling dangerous in any objective way - and in comparison to what. Is it more or less dangerous than common activities like going in and out of a wet tub, walking across a icy parking lot, or walking on a wet linoleum floor?
Cycling is not dangerous. It is not only reasonably safe, it is better for longevity and over all health than not cycling. The "dangerizing" of cycling represented by the popularizing of helmet-as-necessity is bad for overall health and longevity. This is my "opinion" and probably as provable as the opinion that cycling is dangerous.
Cycling is not dangerous. It is not only reasonably safe, it is better for longevity and over all health than not cycling. The "dangerizing" of cycling represented by the popularizing of helmet-as-necessity is bad for overall health and longevity. This is my "opinion" and probably as provable as the opinion that cycling is dangerous.
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02-22-23, 04:48 PM
#575
Full Member
No you didn't. The answers you gave are fairly meaningless, since they include neither your assumptions nor most of the relevant calculations.
As it happens, the specific answer you offered (10.2J over ten seconds) is correct if the added mass is placed at a diameter of 755mm. Which is kind of a strange assumption, as is your use of rotational velocity as an alias for speed at the opening of the problem.
.
As it happens, the specific answer you offered (10.2J over ten seconds) is correct if the added mass is placed at a diameter of 755mm. Which is kind of a strange assumption, as is your use of rotational velocity as an alias for speed at the opening of the problem.
.