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Thread: gear torque/speed

  1. #1 gear torque/speed 
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    Hi, I am a little confused about torques and speeds in a gear train. It is my understanding that you can either increase the torque and decrease the rotational speed using a gear train, or vice versa. Basically torque and angular speed are inversely proportional.

    I have also seen the same spoken about motors. You can either have motor with low torque and high revs/s or vice versa.

    My question is, if the torque is higher, wouldnt the rotational speed also be higher? How do these two quantities go opposite to each other? In a gear train shouldnt the gear at the end rotate faster if the torque is increased thru the train?


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  3. #2  
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    It might be easier to begin with a simple example like a balance. When you have two equal weights, you can balance them by placing them an equal distance from the fulcrum or balance point. Torque applied to the beam is equal for both weights. So torque is a measure of perpendicular force applied a distance from the fulcrum or center of gravity. If you have twice as much weight on one side of the balance, that weight must be moved half the distance to the fulcrum to restore balance. So Torque is force times perpendicular distance. F=m*a and T=m*a*d

    With a wheel or sproket, the fulcrum is the axle and the distance is the radius.

    Now you know that smaller radius sprockets must turn faster to keep up with larger sprockets right? You also know that force must be conserved right? Ok then if you apply the same force to a large sproket (which turns slower relative to the small sprocket) the torque is greater.

    Do you see how that works?

    I will pause here before going on to your other questions.


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  4. #3  
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    Cypress, I appreciate you trying to help. I am actually familiar with moments and moment arms. Maybe I should have elaborated on the situation. Im actually in my senior year of mechanical engineer, and for our senior design project we're using a gear train. We have a fan attached at the end of the gear train. The purpose of the gear train is to increase angular velocity of the fan at the end of the train (so basically big gear with lots of teeth to small gear with few teeth, multiple stages). So angular velocity increases thru the gear train. However this would mean that torque decreases thru the gear train, since in a gear train torque and angular velocity go opposite to each other. So we reduce torque thru the train, and fan spins faster, but has less torque.

    Now someone on the team said, "well, if the fan is supposed to spin faster, you should need more torque, because more torque makes things rotate faster". And that stumped me. He is looking at it from a different perspective. And it seems logical that more torque makes things spin faster. yet my knowledge of gear trains tells me we need more angular velocity (and therefore less torque) to make fan spin faster.

    I am unable to reconcile these two conflicting ideas. Which one seems correct to you and why?
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  5. #4  
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    I'm sorry to have presumed your level of knowledge, garyofcourse. I decided to answer this question because I had to address this issue as a mentor for the high school robotics team a couple years ago. We were using some very high torque low speed drive trains for a rail gun and also had to build a powerful vacuum powered by a very wimpy motor.

    Also in my current work I manage a team of engineers and specialists doing surveillance for some serious rotating equipment. We are watching over some 1500 horsepower 50 stage axial flow pumps 8000 feet below sea level on the ocean floor. We must ensure that we don't break the things because a trip out of the hole is 100 million dollars. We also have some 15,000 hp gas turbines connected to multistage centrifugal compressors that top out at $4 million if they fly apart.

    Ok so let's try this again.

    Recognize that once the fan is up to speed the only torque required is to match the energy required to keep the fan spinning. Given this reality, what you need to do is match the energy demand of the fan with the maximum energy output of the motor. In other words, you need to determine what motor speed gets you the most energy out of the motor (not exactly the same as energy in because motors have variable efficiency curves). This is the ideal motor speed. Then you use this power figure to estimate the fan speed that would consume this amount of power and then design a transmission that sets up this gear ratio.

    The difficulty you are having is that you have let your friend conflate torque and power requirement. Don't make that mistake or you will end up with a motor that is not running at its peak power output and though your gear ratio will be amazing (thus impressing the finest transmission specialist), the motor will either be bogged down or coasting, and the fan speed will be much slower than ideal. Power is the name of the game.

    Edit: Another point is the speed is not directly a component of torque. Torque is required to transfer force but constant speed itself requires no additional force. If all you were doing is spinning a shaft and a zero pitch fan, as I am sure you are aware, torque would not matter much at all. The force (and power) comes in when you realize that the fan is pushing something so that the faster the fan spins the more power you need.

    If you also need the fan to spin up fast, then you need to figure out how fast you want it to accelerate and reserve some power for that. I bet you can do that calculation too.

    edit again: crud I forgot to answer your direct question. You are more correct than your team mate but he/she does have a point. It is true that the torque required as measured on the propeller shaft must increase with the speed of the propeller/fan since power requirement is obviously increasing. This is where your friend is correct. It is also true that as you gear up the drive shaft (fan) relative to the power shaft (motor) available torque transfer from the power shaft to the drive shaft drops as you describe.

    As I explained above, the trick is to match the two systems so that you get the most out of your motor. I assume this is the goal but you didn't say exactly.
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  6. #5  
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    I suggest you think about it in terms of impedance matching, except you will have to "translate" it from the world of electricity to mechanics.

    A gear train is the mechanical analog of a transformer. It "translates" high torque and low rotational speed to low torque and high speed, or vice versa. For greatest power output, the load and the power source need to be adjusted to each other.

    Your fan is probably light and easy to turn, but you have to spin it quickly. If you tried to turn it by hand, you would soon be out of breath, spending more energy on moving your hand itself than the fan. If you do it through the gear train (having replaced the motor with a crank and handle), you will have the feeling that your hand is slowly turning something heavy and with lots of friction.

    The same happens when you change gears on a bicycle. If you ride downhill in low gear, the pedals provide almost no resistance for your feet to work against, and you could just as well be kicking in the air. So you switch into higher gear and the "feel" of the pedals gets better adjusted to the properties of your legs: capable of providing a lot of force at relatively slow speed.

    Hope this helps,
    L.
    Leszek. Pronounced [LEH-sheck]. The wondering Slav.
    History teaches us that we don't learn from history.
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  7. #6  
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    Neglecting friction losses in the gear train, power P remains constant, while torque T and pulsation (probably) change. The following relation is what you need:


    Now, in the case of the fan, it's quite likely your friction losses cannot be neglected. In fact, if you increase speed, you also increase the friction your motor feels. If it's a small fan, nearly all of the power would go to friction losses.

    What kind of motor are you using? Most electrical motors run fast enough for the average fan, and a DC motor might actually benefit more from a reduction.

    EDIT: for clarity, I was referring to the mechanical power, which will also depend on the speed and torque through the speed-torque characteristics of the motor, which vary wildly depending on the type. (and also, as mentioned by other, on the motor efficiency)
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    After thinking abut the way I answered your quedtion, it would have been more accurate to say it this way:

    You are speaking in terms of available torque provided by the power source and delivered through the transmission (a torque divider and speed multiplier).

    Your team mate is speaking in terms of torque demand required by the fan which is a function of the fan power requirement and therefore fan speed.

    The two functions work opposite each other and it can get confusing. when connected of course they will balance. The key once again is to gear the thing so that you use all available power from the motor as I descrbed above.
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  9. #8 think "actual" torque vs "available" tor 
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    Hi guys, thanks for replying. Some of you asked about the motor, well there isnt actually a motor, I just used that word to indicate there is some sort of input. In reality we have a compression spring which is compressed, and then it expands back out. As it expands it turns a cam. And this cam is on the same axil as the first gear of the train. Basically the cam does one revolution as the spring expands out. But the fan needs to do many revolutions for one revolution of the cam. Hence the gear train.

    Cypress, what you said in your last post might be the answer to my question. I am now thinking that in a gear train when we talk about torque, we are talking about "available" torque and not "actual" torque as I earlier thought. My gear train increases angular speed and reduces torque. In classical physics we think that reduced torque on a spinning body will reduce the angular speed. So the last gear in the train having less torque than the first gear should spin slower than the first gear, and thus fan should spin slow. But I guess we are not talking about torque on the last gear, but rather the torque resistance that the last gear can overcome. Since the fan offers little resistance to rotation, the torque it feels is probably a very small torque, much less than the "available" torque of the last gear from the gear train calculations.

    Another way to look at the idea of actual versus available torque - if Im not mistaken, you can get two motors with the same torque but different shaft rotation speeds (basically they have different power). So torque here again is available torque, otherwise the same torque would cause both shafts to spin the same angular speed if it was actual torque applied on the shaft. Come to think of it, if torque on the last gear was actual torque rather than available torque then the last gear in any gear train would be accelerating all the time, or atleast till air resistance forced it to stop.

    Lemme know if you guys agree with this actual versus available torque distinction. As such we wont be changing the gear train either way as we've manufactured the gears and the semester ends in 3 weeks. But I wanted to think this thru, especially since the guy's question stumped me.
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    Yes I agree you have described it correctly.

    BTW, What exactly is the goal? You implied it was to maximize instantaneous angular velocity, is that right?
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  11. #10  
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    Oh the goal wasnt anything complicated. Since we are running the fan off energy stored in the spring the time it runs depends on speed at which it runs since energy is limited. So wed decided we want it to run at speed x so it would run for time t. The gear train was created to make it rotate at this speed x revs/sec.

    The bigger idea behind the project was to make a non electric fan. Person cranks/pulls a rope a few times and that loads up the spring. And then fan runs off that for few mins. Good for off the grid living, or for parts of the world where electricity is not available. Not the kind of million dollar equipment you usually deal with
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  12. #11 Re: think "actual" torque vs "available" 
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    Quote Originally Posted by garyofcourse
    Hi guys, thanks for replying. Some of you asked about the motor, well there isnt actually a motor, I just used that word to indicate there is some sort of input. In reality we have a compression spring which is compressed, and then it expands back out. As it expands it turns a cam. And this cam is on the same axil as the first gear of the train. Basically the cam does one revolution as the spring expands out. But the fan needs to do many revolutions for one revolution of the cam. Hence the gear train.
    That explains a lot. In this case, power isn't constant, but torque is.
    Quote Originally Posted by garyofcourse
    Cypress, what you said in your last post might be the answer to my question. I am now thinking that in a gear train when we talk about torque, we are talking about "available" torque and not "actual" torque as I earlier thought. My gear train increases angular speed and reduces torque. In classical physics we think that reduced torque on a spinning body will reduce the angular speed. So the last gear in the train having less torque than the first gear should spin slower than the first gear, and thus fan should spin slow. But I guess we are not talking about torque on the last gear, but rather the torque resistance that the last gear can overcome. Since the fan offers little resistance to rotation, the torque it feels is probably a very small torque, much less than the "available" torque of the last gear from the gear train calculations.
    I don't completely agree, although it might be that you are expressing yourself confusingly. I'll try to explain.

    You can't say that in "classical physics" (whatever that is) reduced torque means reduced angular speed for two reasons:
    1) torque will at first approximation influence acceleration, and not speed. In your case, a lot of the torque might go to accelerating the gears, at least at first.
    2) your last gear is not an isolated system. You have two torques on it, one from the previous gear, and one of the load. The difference between both is acceleration of the gear and friction. The fan also has two torques acting on it, one from the last gear, and one from the load (air resistance). The difference between both is acceleration.
    A gear train is a geometrically locked system, so you can't look at the separate parts to study the dynamics. A small torque at the end is still felt as a large torque at the actuator side, which is the source of the rotation.

    All torque in the system is actual torque. Some goes to accelerations, some to friction and some to air resistance. This is different from an electric motor for example, where the torque changes with speed, and there is a certain speed at which the torque is maximum, which could be called "available" torque.

    Quote Originally Posted by garyofcourse
    Another way to look at the idea of actual versus available torque - if Im not mistaken, you can get two motors with the same torque but different shaft rotation speeds (basically they have different power). So torque here again is available torque, otherwise the same torque would cause both shafts to spin the same angular speed if it was actual torque applied on the shaft. Come to think of it, if torque on the last gear was actual torque rather than available torque then the last gear in any gear train would be accelerating all the time, or atleast till air resistance forced it to stop.
    For motors this is slightly more correct, depending on the type. Most AC motors are designed for a certain rotational speed, and they will only offer the torque required of the application, or stall if the load is too high. Some motors, such as an asynchronous induction motor will run slightly slower as the load increases.
    For most DC motors, however, the speed depends a lot on the load, and you can't really talk about two motors with the same torque but different rotational speeds, without specifying the load (they have a certain maximum speed and a maximum torque, but those don't apply in the same point of application). This is also true for your spring system, in which the last gear will indeed be accelerating all the time, or at least until air resistance and friction force it to stop (accelerating).
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