Class 1 lever, pulley 2:1 ratio. You know physics? What motor and gearbox i need?

*Thanks to everyone in advance for helping. To the people that have poor memory retention and can't understand my entire post after reading, please refrain from answering. To everyone that are just here to size me up and tell me what i should or shouldn't do based on the skill level you think i have, please refrain from answering.

Hi everyone. I am new here and i started to design my own 3dof simulator a few weeks ago with heave motion. I am following someone else's tutorial on an other website with a lot of detailed information that seems to have been worked out by a mechanical engineer with knowledge of physics. I am trying to take all that base knowledge and reinvent my own with my personal tweaks and mechanical systems. I have learned about the 3 classes of levers in physics, i have looked into gear ratios and pulley systems to learn as much as possible about everything and the science about it. The only problem is i'm not a scientist so no matter how much i learn, i don't know where to look to know how to combine and apply all this knowledge to my own design. Maybe for some of you, when you look at my drawing, you will instantly think its a piece of cake, well, that's good because you are the kind of people i'm reaching out to. It will be a 3dof rated at 200kg so 200kg divided by 3 is 67kg for each motors. 67kg is roughly 650 Newton but i round it up to 670 to make sure my choice of motor is powerful enough. 67kg is a force of gravity of 670N and it will be applied to a lever of 150mm with a range of 160 degree of revolution and that lever is attached to a 200mm pulley. By my calculations, the 200mm pulley will be pulling 502.5N but i could be wrong, i'm not sure. The 200mm pulley will be linked to a 100mm pulley with a timing belt. the 200mm pulley and 100mm pulley have a 2:1 ratio but i'm not sure it really applies here since i know ratios work with gears but i don't know if its the same with pulleys. So if there is a ratio, it would probably mean the 100mm gear would be receiving a pull of half of 502.5N? In the drawing i used a 20Nm motor with a rpm of 180 but its just an exemple and i have no idea if that will be enough. lets say the motor is 20Nm so it means the 100mm pulley can pull 400N? if the gear ratio works on pulley and i have a 2:1 ratio, does it mean my 100mm pulley will be pulling twice 400N on the 200mm pulley? As you can see here, i'm really confused. How does it even work to calculate the torque and rpm of 2 pulleys that are working against each other? There is no such teaching on youtube, all they show is simple pulley and gear system but nothing explaining what to in the case of 2 opposing forces, combined with a class 1 lever. So to help you help me, i absolutely need to have a 150mm lever receiving the 670N of force and those a 2 numbers i can't change. What pulley sizes and motor torque and rpm i should shoot for? Please don't tell me i need a 100Nm motor with the lever and no pulleys. I already know that. I'm trying to make an efficient but cheap system so by introducing pulleys i could get a weaker motor that could still do the job. Motors and PSUs are the most expensive parts and in order to get cheaper motors and gearboxes i need pulleys. I would hope to be able to get away with a 1Nm servo with a 20:1 gearbox. Do you think its possible?

Thank you for your time

Thanks for your help noorbeast. I will take a look at the link, there seem to have some interesting information when i just glanced at it. I know the weight wont be perfectly dristributin on the 3 motors but the spring loaded shaft in the middle of the platform and centre of the gravity will be pretty much centered to the 3 motors and the 3 motors will all be at equal distance so more or less, it will be even enough if i pick a motor that can handle at least 200kg together plus more. I wont take a motor that is just the bare minimum, i'll pick something at least .5x stronger than the bare minimum. I just need to know what the bare minimum would be so i have something to base my decision from. Thanks again for your help, its always very welcome

Sorry everyone, i think i just figured it all out on my own tonight while i was watching more videos on the physics of pulleys. My problem is that i was visualizing the 2 opposing forces clashing each other and fighting as who was gonna be taking advantage of the 2:1 ratio before the other and i was thinking whatever is the strongest force was gonna win but of course that's not how it works and i was just really confused and needed help to clear it up. Here's what i came to understand. I need to calculate the force of 1 side all the way to the other side to see what the other side need to be an equal force. I'll explain:

We need to know, what will be the Torque generated by the 670N down the centre of the 100mm pulley. That will tell us how strong the motor need to be to counteract it:

Torque applied at the centre of the 200mm pulley: 670N x 0.15m = 100.5Nm
Now with the torque of the 200mm pulley we can find the force it generates on the outer of the 200mm pulley: 100.5Nm ÷ 0.2m = 502.5N
We know that the 2:1 ratio between the 200mm pulley and the 100mm pulley is cutting the torque from the 200mm pulley in half: 502.5N ÷ 2 = 251,25N
Now that we know what is the force applied on the outer of the 100mm pulley, we need to know what is the torque of that force down in the middle of the 100mm pulley: 251.25N x 0.05m = 12.5625Nm

There it is: The minimum torque necessary on the 100mm pulley that will counteract the force generated by the weight of the cockpit of 200kg on 1 of the 3 motors is 12.5626. We know there is a loss in gears and pulleys system so its actually less than 12.5625Nm but its pretty irrelevant at this point.

Now that we know a Servo of 1Nm with a gearbox of 20:1 would be 20Nm, we could calculate the loss that gearbox will incure to the total Nm output of the motor and to know exactly what would be the final force pushed by the motor to the lever, we could find a loss calculation for pulleys ratio but i'm pretty sure it would be overkill. I think to be safe, a 1Nm servo with a 25:1 gearbox would do just fine.

If i made any mistakes and got something wrong or i just simply understood it all wrong and i don't get it at all, just let me know and explain why and what is the correct answer.

If i'm right, i hope i helped someone and good luck with your creation.

If i was using a motor mounted directly to the shaft, i would need a 100Nm motor. I already said that. Do you have any idea how would that cost? Yes, an arm and a leg. By the way, i'm not talking about a regular DC motor. I need a motor that can hold a position. DC motors can push hard but they can't hold a position resulting the rig to collapse every time there is no signal. Only a servo motor or stepper motor can hold a position at relatively high torque. Also you keep bringing up the loss like it counts for 90% of the whole thing and i think you're blowing it out of proportion. I really believe that by using the pulley system combined with the class 1 lever i am not just doubling my motor torque and dividing the weight in half. I'm just applying a few simple physics principles together to gain a mechanical advantage and again, i think you're blowing out of proportion that whole loss thing and you are dumbing down the whole pulley system to a point, i wonder you really understand how it works. I looked at the charts for loss and its not that bad. With the money i save on the steppers i could get a decent 20:1 to 50:1 gearbox and the loss would be negligeable. If really my math is wrong, just prove it, don't come here with an other unrelated math problem and act like it answers mine.

Moderator EDIT: Personal attacks are not allowed. Feel free to argue the concepts, just don't make it personal.

It would appear i made a mistake in my equations.

When i was at step 2 that goes like this:

"Now with the torque of the 200mm pulley we can find the force it generates on the outer of the 200mm pulley: 100.5Nm ÷ 0.2m = 502.5N"

i used 0.2m as the divider of the Torque generated by the weight applied on the lever. 0.2m is actually the diameter of the 200mm pulley and its wrong. I have to use the radius which is 0.1m. The 150mm lever and the 200mm pulley act as a class 1 lever. In this case, the center of the 200mm is the fulcrum. Therefore the class 1 lever is composed of the 150mm lever and the radius of the 200mm pulley.

The correct equation is then: 100.5Nm ÷ 0.1m = 1005N

So now that we know the forced delivered to the 100mm pulley from the 200mm pulley is 1005N and NOT 502.5N, we can take in account the 2:1 gear ratio and do 1005N ÷ 2 = 502.5N

From there we can keep following the same steps i used before and find the torque applied to the center of the 100mm pulley this way: 502.5N x 0.05m = 25.125Nm

Now its pretty clear that the Minimum Torque required from a motor to counteract the force of the cockpit weight and opposite it, is something over 25.125Nm. The ideal would be a servo or step motor of a holding torque of 1Nm and a gearbox of something like 50:1 or more. Luckily, i found stepper motors of around 4Nm for less than 60$ and a gearbox of 20:1 would be largely sufficient.

Now noorbeast, i really wonder what the hell you contributed to my problem and how you can imagine that your link backed anything up to support with my system. You just seem to be arguing that whatever your system is(which is obviously not in favour of mine), is better than mine. If you want to spend 2000$ for 3 motors, knock yourself out.

I still have to calculate the rpm of a stepper motor to achieve 160 degree of revolution on the lever within 1 second. It means the 200mm pulley would need to achieve a full 360 degree revolution in 2.25 seconds. The rpm of that 200mm pulley would then be 26.67 rpm. It doesn't sound like much. Again, gear ratio of 2:1, the motor should have a minimum rpm of 53.34 rpm. I'm pretty sure that's manageable. Ok, with this post, i hope to help people so i'll give the formula clear and simple for the people trying to figure out their own.

in 1 full revolution there is 360 degrees
i need in my example to make 160 degrees in 1 second
360 ÷ 160 = 2.25 so thats 2.25 second to achieve 360 degrees
in a minute there is 60 seconds so 60 ÷ 2.25 = 26.67 rpm
then the gear ratio of 2:1, big wheel to small wheel, it always go faster so obviously the gear ratio will multiply, by 2 in my case for a total of 53.34 rpm

for anyone that wish to use pulleys and learn gear ratios and how to calculate them when its a bit more complicated that just 2:1 like in my case, there is tons of information on the internet but Youtube is always the best place to learn

Great way to get friendly help here from more experienced people than you! But you seem to think you know it all already anyway so why are you even here asking? There are several names for people like you, but I refrain! Probably best if you just ask your questions on YouTube from now on!

@Onsc3ne I strongly dislike the way you've been treating one of the most helpful and experienced members of this forum, but I'm going to try to point you in the right direction.

You fail to account for motor power output (Power = Torque x Angular Velocity). A motor delivers maximum torque at stall speed. It delivers no torque at maximum speed. A motor that has sufficient torque at stall and meets your speed requirements at no load will not be sufficient when you put a load on it. You need to find a motor that meets your power requirements, then use whatever mechanical means you wish to meet your speed and torque requirements. Reducing the required torque does not reduce the size or cost of the motor you'll need.

You are not only holding up the mass that is placed on your simulator against the force of gravity, but you also need enough power to accelerate it (Acceleration = Force / Mass). The load is not going to accelerate linearly, meaning you'll need to watch a lot of YouTube videos about calculus if you wish to solve your power requirements mathematically. It's a lot easier to be kind to those who have already solved this by trial and error and learn from their experiences!

Hi @Onsc3ne, you are forgetting that know one here owes you anything and rudeness will get you nowhere. You keep realizing that you have made a mistake, or missed something in your calculation, it is not just in your calculations - good luck with your build. .

Short answer is no.

You will sacrifice speed for mechanical advantage. A motion simulator needs to be fast and precise, both of which will need a fast (100 - 200 rpm) yet powerful (150 - 300W) motor. There is a reason that almost all simulators you see in this forum or elsewhere have got either wiper motors (40W - 60W) or motors that are close to, even exceeding 200W. Yes you could probably "imagine" an ideal motor in mathematical terms but at the end of the day you have to find what you can actually purchase and what the cost would be to modify that system. Usually 9 times out of 10 it's far easier and cheaper just to buy a big ass motor.

Moderator EDIT: Personal attacks are not allowed. Feel free to argue the concepts, just don't make it personal.

http://www.blocklayer.com/pulley-belteng.aspx

@Onsc3ne I have no idea what I have just read ???? And to be honest I forgot most of what you just said something about refraining from posting but my memory is terrible and you know what I've weighed you up and I won't be posting anything on this thread........ can't wait for @bsft 's comment

thats why I didnt reply, I have got no idea either @SilentChill

Take it easy buddy ...

Come on guys ... He is really cute .

Sorry if you don't like the fact that i try to keep my post on point and clean from branching comments that deviate from my question you can read in the title. I would expect from an experienced person to have the ability to attack a question buy bringing a related answer instead of bring an answer that raise many questions. There is a very specific name for people like you but i will refrain. Thanks for contributing to a post that could have helped many by adding your so useful 2 cents of wisdom.

You know what? I don't know who's experienced here or not and i don't measure how much respect someone deserve by the reputation they built in the past within a group i don't even know. I measure the respect by how someone is genuinely trying to help with actual information that adds to a question that was already comprehensively delivered. If someone just want to pat themselves on the back by throwing a piece of dry bread, i couldn't really care less who they are in an other life. That said, i appreciate that you realized that these posts are made public and you decided to add something actually useful that will not only help me but anyone tinkering with a similar concept. You're the only bigger man in the room so far.

I could very well stop being greedy and hire a mechanical engineer and never start a post with drawing and explaination and people like Ben V adding actual useful information for everyone to leech for free. You are free to just not come here if you think the people asking questions are only taking and not giving in return.