Hello in the context of my project of teminale to model a gear system transforming 5 rpm in 2000rpm. I am currently encountering a problem since I have, to model this system I have multiplied large on small, small on large and so on. But I'm not at all sure. I attach the photo of my gear system if you want to give me advice or other. Thank you in advance for your answers!
Hello
You have to imagine the gear system like that of a bicycle.
r= driving wheel / driven wheel.
If the pedal plate is larger than the wheel platter --> the speed is increased and the torque decreases because more power is needed to pedal
conversely, the chainring is smaller at the pedals and larger at the wheel --> the speed decreases and the torque increases because less power is needed to pedal
V= 2000 x (50/100) x (20/100) x (30/80) = 75 rpm
Reminder: V = W x R N = (30 x W) / Pi P = C x W
Edit: dsl not awake!
Hello?
Forced to go through a train of engenage?
Can we go through an epicyclic train? For such a reduction ratio it could be useful.
https://fr.wikipedia.org/wiki/Train_%C3%A9picyclo%C3%AFdal
Fred
Going from 2000 rpm to 5 rpm is a big reduction.
Your basic idea is classic, except that you need to review your values (ratio = leader/led).
Also look at epicyclic trains.
Especially since it's 5 at the start and 2000 at the end, it's going to take a couple to start all this!!
Ok so that means that my calculations are not good at all! Just I have to start from 5 to have 2000 and the beginning of my system and the sprocket of 80 so it gives me V = 5 x (80/30) x (30/100) x (100/20)x (20/100) x (50/100) = 2.6 x 0.3 x 5 x 0.2 x 2 = 7.8 rpm do we agree? It doesn't multiply anything at all!
I also thought about the epicyclic gear train but I don't know too much about it so I preferred to compose with a simple gear train. Afterwards, if you have something to suggest to me, we could see it together but I have a big emultation.
On the picture I only see 3 stages of reduction (80/30 - 100/20 - 100/50) can you tell me where are the other 2 you see?
Maybe a worm gear reducer would be more suitable... but expensive...
R = Nbr of threads for the screw / Nbr of teeth for the pinion
If it's an electric motor, pk don't use a variator?
Hello
If your entry sprocket at 5 rpm is the 80 tooth one then:
Voutput = 5 x (80/30) x (100/20) x (100/50) = 133.33 RPM for 50 tooth sprocket.
Your 30-tooth sprocket and the first 100-tooth sprocket rotate at the same speed.
Your 20-tooth sprocket and the second 100-tooth sprocket rotate at the same speed.
To get an output speed of 2000 rpm, you can for example:
Voutput = 5 x (80/20) x (100/20) x (100/20) x (100/25) = 2000 rpm, which adds a floor to you.
Kind regards
Definitely, I go back to bed, I took the exit gear instead of the entry one...; ( ?
On the other hand, suddenly the resistive torque must be monstrous in that sense...
In general, the power does not vary, so if the speed decreases the torque increases (P = C x W)
The best way to deliver such a large torque is through an endless screw.
Following a last minute modification my colleague takes a less powerful engine running at 120 rpm. The joys of working in a group!
But here I have too many different opinions on the subject, I don't know what to do, I'm lost!
Is it a stepper motor? How do you get such a small speed (120 rpm)?
Power (PU)? / Speed in? / Speed out? / Special conditions (friction, inertia, ...)?
On installations requiring significant effort, it is advisable to install a mechanical fuse (belt, coupling, etc.) in sensitive areas to avoid damage to the system.
If you're interested, I have to transform fitness machines already provided so that they can produce electricity that will be used to feed from the streetlights of the fitness area wholesale. Thanks to the rotation angle of my machine and the number of repetitions I found for my first machine 5rpm. Since my colleague changed the motor, I have to adapt it via a gear system with a multiplication cf of a little more than 24. The new motor makes 20 W so a torque of 0.1 if I'm not mistaken. I thought I would take 2% loss per gear because calculating the losses seemed very complicated to me and it didn't even bother me.
Hello
What type of lighting is it for? And for what voltage (volts)?
Your project is nice but I doubt very much that it will be profitable.
I think you would be quicker to use an alternator or dynamo to charge batteries managed by a control system. In particular, the speed will not be constant. In addition, you could integrate this kind of installation with other fitness machines. I think if the machine runs all day and part of the evening, the charging time might be enough to power a lighting area. It would even be more interesting to use a 12vdc power supply network for lighting.
If it's to have 230v, it will be enough to use a 12vdc / 230vac transformer, but there will be significant losses.
I don't really understand the concept of your machine, physical effort creates an oscillatory movement? If this is the case, two relays would have to be integrated to make an electromagnetic lock driven by the rocking movement, so as to make a polarity reversal for a battery.