Keep doing these. You will always get a mechanical engineer to chime in and tell you how wrong you are; it doesn't matter. Your presentation format makes it easy to understand the underlying concepts of riding. Your subscribers appreciate your efforts so thanks again and keep going!
The force you calculate is the average force applied over one revolution of the cranks. The peak value for one leg would be significantly higher (somewhere in the downstroke) and much lower somewhere near the end of the upstroke I guess. I think that the Garmin Vector pedals can give you the forces over the crank angle as part of their cycling dynamics features. Have you ever tested those?
You were saying that the torque needed to push 300w was approximately like lifting 19 kg. With that in mind would it be accurate to say that squatting 19 kg, 90 reps in one minute would be equivalent to pushing 300w @ 90 rpm for 1 minute?
how tall are you and what crank length are you on? im 175cm and im on 172,5 i want to get shorter but im not sure how short, 170 or then all the way to 165 seems to be the options.
I measured femur to around 47cm and inseam to 84 cm. when im running for example i favor higher cadence and shorter steps, its not the same thing but i feel like maybe its something to it. it just sucks its allot money to get new cranks to try.
Not quite right on this one. You were talking about the power (generated by ciclist) and in the equation is the forced produced by cyclist multiplied by it's speed. It doesn't have anything to do with an external system for de resistance forces . You can generate 200w at a speed of 8.33 m/s (30 km/h) by using a force of 24,009 N. To generate the same wattage but at a speed of 9.72 m/s (35 km/h) you need to use a force of 20.576 N . I picked up 2 situations depending of the speed you are traveling at ... Just when I thought I got that video that I was asking for :)) But , keep on doing em
Vlad Biciclistul wrong. If you want to travel at a constant speed, your power needs to be equal to resistance forces×velocity. The power you generate is Tx2xPIxCad.
Vlad Biciclistul noone said it can. If you have set drag forces at a speed, multiplied by the same speed, you get the power required to travel at that constant speed, given that the resisting forces don't change. The power meter measures torque and cadence, as described in the video.
Hello again. This time you're talking about things I actually know a lot about :) [check out my publications]. Sorry, but there are a lot of inaccuracies, and you're not really talking about the important points, and like everyone else in the cycling world you're missing an extremely important point (point 10 below): 1. You say power is the rate of change of energy, but that's with respect to time (you only say that later). The rate of change of potential energy with respect to distance is force. As you do in the rest of your video, there is no reason to introduce derivatives. 2. You use the term "velocity" incorrectly. What you are really talking about is "speed," which is a scalar and the magnitude of velocity. In particular, the angular velocity of the pedals depends on how fast the bicycle is turning, but that quantity is not taken into account in any of your considerations. You should stop talking about velocity in this context. 3. Power is the ability to do work quickly. Two people with the same weight cycling up a 1000m hill do the same work against gravity, the one who does it twice as fast has double the power. Since cycling is about coming in first, this explains why power is so important in cycle racing. 4. A person with half the weight does half the work against gravity going up that same hill, so to have the same time as the others, he only needs half the power. This explains why power to weight ratio is so important. 5. Similarly, on flat ground, one expends energy against wind resistance, so the important quantity is the power to aerodynamic drag ratio. 6. You need to exert power to climb hills and overcome aerodynamic drag and you apply that power through the cranks. The instantaneous power you apply is simply the force you apply to the pedal time the pedaling cadence times the crank length. 7. In general, you ignore crank length since you compare power using the same cranks. So power is essentially force * cadence. Pedal based power meters require you to specify the crank length. 8. Strength is the application of force. The lower the cadence, the more power is generated by force (a rider who pedals slowly and goes fast is strong). It follows that great strength is most valuable for standing starts in a big gear and the first few pedal strokes after that. 60kg cyclists with little discernible muscle have ridden 1:03 standing kilometers, they lost the 2 seconds compared to 85kg cyclist in the first few pedal strokes. 9. The universally accepted statement is that to go faster for similar bicycle and rider, you should increase the force on the pedals or increase the cadence. 10. This is not correct. You can also increase power by doing more work per pedal revolution while keeping the same cadence by having a force profile which is not always greater than a slower force profile but which has greater area under it, or by eliminating resisting force in one leg when the other is pedaling. It is not at all obvious how this comes about and whether it can be learned. Enjoy! -ilan
On number 3 You say that a rider at 50kg would have to produce only half the amount of watts as a rider at 100kg to climb up the same hill at the same time, thus having to do half the work. This is not true if the 100 kg rider's FTP w/kg is equal to the 50 kg rider's FTP w/kg. So the 100kg rider would be riding at the same amount of exertion in relation to their own bodies capacity as the 50 kg rider. Not twice as much.
I didn´t mean to, because they are pointless, but if you insist: 1.) I wrote dW/dt, I think it is pretty obvious that it is a time derivative. What does force and Ep even have to do with this? 2.) it is scalar and it is correct, stop splitting non existent language hairs 3.) the whole video is about this 4, 5, 6, 8, 9 ) doesn´t work like this at all in real life, plus explaining this was not the point of the video 7.) Force times cadence yields a unit of N/s or N/min. The unit of power is J/s so this is plain wrong 10.) Doing that will increase the physiological cost just as much as increasing the cadence or the force applied to the pedal. You can´t just gain power for free by pedalling "smoothly".
Isn't F=dV/dx something we tend to use only when the force field is conservative? Otherwise it's kinda hard to fabricate a scalar potential for the force in the first place.
As an engineer. I think you did a good job to explain something that is difficult to explain to the masses. Good job.
Hambini thank you!
Hambini both you guys are awesome 😎
Keep doing these. You will always get a mechanical engineer to chime in and tell you how wrong you are; it doesn't matter. Your presentation format makes it easy to understand the underlying concepts of riding. Your subscribers appreciate your efforts so thanks again and keep going!
Super Strada thanks for the support. On a side note, I am finishing my ME studies in 2 weeks.
Thanks Rony! Super interesting. I'm looking forward to your follow up vids on this topic.
Pian0Mon thanks! I am planning to do one every week.
The force you calculate is the average force applied over one revolution of the cranks. The peak value for one leg would be significantly higher (somewhere in the downstroke) and much lower somewhere near the end of the upstroke I guess.
I think that the Garmin Vector pedals can give you the forces over the crank angle as part of their cycling dynamics features. Have you ever tested those?
OT. What are the benefits of cycling short but steep climbs?
AL Francis that depends on the effort that you produce.
i love this series. It seems kind of useless, but it warms my engineering heart. Looking forward to the rest.
Remi Fjelldal thanks for the support
Fascinating thanks for your concise explanation
simon russell thanks!
Man you totally set my brain upside down, since it's been such a long time that I've done science. 😂😂😂. Great video my man, really educational👍
Hessel Glotzbach thanks!
You were saying that the torque needed to push 300w was approximately like lifting 19 kg. With that in mind would it be accurate to say that squatting 19 kg, 90 reps in one minute would be equivalent to pushing 300w @ 90 rpm for 1 minute?
Liam Sangaku exactly
Liam Sangaku Wouldnt you need more energie to squat 19kg? Because if you squat, you need to squat your own Bodyweight too, or am I wrong?
Der Übermensch You have to push your body weight on the bike as well.
Liam Sangaku Yes but the Bodyweight is variable, but the 300 Watts output not.
Very good, Rony
Thurston Howell III thanks!
Assuming crank length is 172.5mm right?
Frank van den Broek 0.17m, but the differences are tiny
how tall are you and what crank length are you on? im 175cm and im on 172,5 i want to get shorter but im not sure how short, 170 or then all the way to 165 seems to be the options.
Your crank length as more to do with your femur length and overall leg length. Not your height.
visse89 172cm. 165mm on the TT bike, 170 on road and CX. For me it's about getting a good position
I measured femur to around 47cm and inseam to 84 cm. when im running for example i favor higher cadence and shorter steps, its not the same thing but i feel like maybe its something to it. it just sucks its allot money to get new cranks to try.
170 could be a good starting point
Interesting and very basic physics. What are you studying btw?
Raphael Tiziani motor vehicle construction
Not quite right on this one. You were talking about the power (generated by ciclist) and in the equation is the forced produced by cyclist multiplied by it's speed. It doesn't have anything to do with an external system for de resistance forces . You can generate 200w at a speed of 8.33 m/s (30 km/h) by using a force of 24,009 N. To generate the same wattage but at a speed of 9.72 m/s (35 km/h) you need to use a force of 20.576 N . I picked up 2 situations depending of the speed you are traveling at ... Just when I thought I got that video that I was asking for :)) But , keep on doing em
Vlad Biciclistul wrong. If you want to travel at a constant speed, your power needs to be equal to resistance forces×velocity. The power you generate is Tx2xPIxCad.
Ronald Kuba it's a nonsense what you say...
Vlad Biciclistul I guess I'll have to tell all of my professors to return their diplommas. You clearly have no idea how drag forces work.
If you tell me how on Earth can a PowerMeter measure any drag force ...
Vlad Biciclistul noone said it can. If you have set drag forces at a speed, multiplied by the same speed, you get the power required to travel at that constant speed, given that the resisting forces don't change. The power meter measures torque and cadence, as described in the video.
Being an engineer was easier to understand, was educational. Keep up (:
Mohammed Azharuddin thank you!
Hello again. This time you're talking about things I actually know a lot about :) [check out my publications]. Sorry, but there are a lot of inaccuracies, and you're not really talking about the important points, and like everyone else in the cycling world you're missing an extremely important point (point 10 below):
1. You say power is the rate of change of energy, but that's with respect to time (you only say that later). The rate of change of potential energy with respect to distance is force. As you do in the rest of your video, there is no reason to introduce derivatives.
2. You use the term "velocity" incorrectly. What you are really talking about is "speed," which is a scalar and the magnitude of velocity. In particular, the angular velocity of the pedals depends on how fast the bicycle is turning, but that quantity is not taken into account in any of your considerations. You should stop talking about velocity in this context.
3. Power is the ability to do work quickly. Two people with the same weight cycling up a 1000m hill do the same work against gravity, the one who does it twice as fast has double the power. Since cycling is about coming in first, this explains why power is so important in cycle racing.
4. A person with half the weight does half the work against gravity going up that same hill, so to have the same time as the others, he only needs half the power. This explains why power to weight ratio is so important.
5. Similarly, on flat ground, one expends energy against wind resistance, so the important quantity is the power to aerodynamic drag ratio.
6. You need to exert power to climb hills and overcome aerodynamic drag and you apply that power through the cranks. The instantaneous power you apply is simply the force you apply to the pedal time the pedaling cadence times the crank length.
7. In general, you ignore crank length since you compare power using the same cranks. So power is essentially force * cadence. Pedal based power meters require you to specify the crank length.
8. Strength is the application of force. The lower the cadence, the more power is generated by force (a rider who pedals slowly and goes fast is strong). It follows that great strength is most valuable for standing starts in a big gear and the first few pedal strokes after that. 60kg cyclists with little discernible muscle have ridden 1:03 standing kilometers, they lost the 2 seconds compared to 85kg cyclist in the first few pedal strokes.
9. The universally accepted statement is that to go faster for similar bicycle and rider, you should increase the force on the pedals or increase the cadence.
10. This is not correct. You can also increase power by doing more work per pedal revolution while keeping the same cadence by having a force profile which is not always greater than a slower force profile but which has greater area under it, or by eliminating resisting force in one leg when the other is pedaling. It is not at all obvious how this comes about and whether it can be learned.
Enjoy!
-ilan
On number 3 You say that a rider at 50kg would have to produce only half the amount of watts as a rider at 100kg to climb up the same hill at the same time, thus having to do half the work. This is not true if the 100 kg rider's FTP w/kg is equal to the 50 kg rider's FTP w/kg. So the 100kg rider would be riding at the same amount of exertion in relation to their own bodies capacity as the 50 kg rider. Not twice as much.
Ilan Vardi as I said in the video, I will talk about the resisting forces in the next video
Yeah, that really addresses my comments.
I didn´t mean to, because they are pointless, but if you insist:
1.) I wrote dW/dt, I think it is pretty obvious that it is a time derivative. What does force and Ep even have to do with this?
2.) it is scalar and it is correct, stop splitting non existent language hairs
3.) the whole video is about this
4, 5, 6, 8, 9 ) doesn´t work like this at all in real life, plus explaining this was not the point of the video
7.) Force times cadence yields a unit of N/s or N/min. The unit of power is J/s so this is plain wrong
10.) Doing that will increase the physiological cost just as much as increasing the cadence or the force applied to the pedal. You can´t just gain power for free by pedalling "smoothly".
Isn't F=dV/dx something we tend to use only when the force field is conservative? Otherwise it's kinda hard to fabricate a scalar potential for the force in the first place.
V good video !!!!
Klaudiusz Zieba thanks!