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Automotive How can torques be summed algebrally?

  1. Apr 12, 2017 #1
    Everytime I hear some informations in videos like "......car has 700Nm torque from ICE and additional torque of 400Nm from electric motor so total torque 1100 Nm"

    I cant understand how torques can be summed algebrally. My teacher said if the design is clever enough it can be done and said somethings about differential calculus but I didnt understand.
     
  2. jcsd
  3. Apr 12, 2017 #2
    ?????????

    Why would you not be able to sum them up?
     
  4. Apr 12, 2017 #3
    I can't imagine because when a rod spins under a force if you want to increase the torque you should apply more force and the force you will apply must be larger than the first force. But again, if you apply a larger force new torque value will be depended on only the second force not both first and second force. That's why I can't understand.
     
  5. Apr 12, 2017 #4
    I believe that what you are looking for lies in the compounding gearbox. With two different inputs there also needs to be a balance placed in the path. Choosing numbers from your example and adding a couple invented for demonstration. Lets assume an ICE producing 700 Nm at 5000 RPM. and an electric motor producing 400Nm at 2000 RPM. For desired speed we need an output shaft at 1000 rpm. In a simple system the ICE would be operated at a 5:1 ratio and all is good, or the electric is run at 2:1 the same result is achieved. However if the electric and the ICE are coupled at the input they will interfere with each other and often a reduction in force is seen.
    If one creates an equalization that allows both to operate at their own speed and deliver to an input to the transmission they will both contribute. This is highly simplified as the delivery curves of an electric and ICE are quite different and there needs to be some supervisory control input. Also consider the volumetric changes of the combustion engine. The listed number is only at WOT at a particular RPM. As the throttles close the volumetric efficiency goes down and the output is reduced. A similar thing happens if one artificially increases the rpm of the engine (with say an electric motor) above where its normal operating state the efficiency changes usually reducing output. These in most cases have the ICE forfeiting power to coexist with the electric.
    You are right that the larger force will tend to take over the load however with gear train and control input one can simply sum the outputs.
     
  6. Apr 13, 2017 #5
    Firstly a for spins under a moment, the net torque will dictate how it spins.

    Why does a second forces have to be larger?

    If I attach a chain to a wall and tell you to pull on it. If I then pull on the same rope obviously we do a better job of pulling down the wall.

    If I attach a second rope and pull harder than you, does that make your efforts 0?

    Force is a vector. Torque is also a vector.

    Google: Torque vector, torque right hand rule.
     
  7. Apr 13, 2017 #6
    Yes , that's what I exactly think. I you attach a spring to wheel and pull harder than me actually in order to pull harder than me you must exert more force than me thats my point, if the both force and torque are vector (in principle yes its true) you should be able to increase the torque without not exerting more force than me.
     
  8. Apr 13, 2017 #7

    CWatters

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    The design doesn't have to be "clever". If two forces are applied to an object the net force is the sum of the two. Its exactly the same with torque.
     
  9. Apr 13, 2017 #8

    Randy Beikmann

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    Building on that, if you mount an engine and an electric motor/generator (M/G) on the same shaft, the engine produces 400 ft-lb torque, and the electric motor 300 ft-lb, the result is 700 ft-lb on the shaft.

    If the M/G is producing -300 ft-lb, the result is 100 ft-lb on the shaft. The M/G is now acting as a generator, charging the battery and/or driving another motor.
     
  10. Apr 13, 2017 #9

    jack action

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    Try lifting a garage door with no springs or counterweights attached to it. Then connect a spring or counterweight to the mechanism. You will notice that you need less force to open the door. The spring or counterweight is applying part of the required force, so a lesser effort is required from you. The sum of both forces is the force you applied when you opened the door without any spring or counterweight.

    If your theory was true, how could you determine which force is actually doing the work? If we push together on a box and I apply more force than you, then I would be doing all the work and you would do nothing? If you began to push harder, then as soon as you push harder than me - instantly - you would be doing all the work and I would be doing nothing? The reality is that even if we don't push with the same force (think an adult and a kid), we will both get tired because we both did work.

    Don't mix speed with force. If you can't keep up with me while I'm pushing the box, then you're not applying any force. You are just running behind the box.
     
  11. Apr 13, 2017 #10
    Yes, you've understood me completely thank you!. Indeed, I'm not arguing the forces because I do know that doesn't make sense as well but I guess I'm confused about the speed.

    I want to know that when you attach an electric motor to the drive shaft, will the system work? Can you apply both ICE's and electric motor's torques on wheels. If it can, then what should be the relationship of speed? Maybe that's why the design should be clever.
     
  12. Apr 13, 2017 #11

    jack action

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    On any type of motor, the torque is continuous and related to rpm. So if both motors are connected to the same shaft, then they have the same rpm and each one has a corresponding torque associated with that rpm.

    Let's assume both motors are connected to the shaft via gears. It is technically possible that one gear set is "floating" between 2 teeth (due to backlash). In that case, only the torque from the other motor would apply. But that situation cannot last long as all the torque from the "free" motor would be used to accelerate its rotating components, which means it would catch up the other one very quickly.
     
  13. Apr 13, 2017 #12
    You can sum them vectorially just like forces and use parallelogram and triangle methods to find the resultant. For notation you identify torque vectors with a double headed arrow.
     
  14. Apr 14, 2017 #13

    Randy Beikmann

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    If you're having difficulty separating torque and speed effects, examine the system without speed! Do a static analysis of the torques applied to a system like a hybrid transmission, and you don't have to worry about any work going into accelerating different components of the system. Calculate the reaction torque at the output shaft of the transmission, or whatever it is that you're interested in.

    After that makes sense, imagine the components in the system rotating at a constant speed. This (quasi-static) case isn't that much different than the static case, except that there is power put into (by the engines and motors) and taken out of the system (by driving a load) which, neglecting friction, nets to zero. But still no acceleration, so that's not a concern.

    These things need to be understood step by step, or you will only confuse yourself (as I have learned myself).
     
  15. Apr 14, 2017 #14

    CWatters

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    This is where real world issues raise their head. You clearly can connect both an IC and electric motor to a cars wheels as that's exactly what happens in a hybrid F1 car. Both ICE and electric motors have performance curves rather than constant performance. They have to be "matched" together and to the car just as an IC engine had to be matched to the vehicle and it's required performance via the gearbox. I'm no expert but I imagine it's pretty complex to get this right, particularly for something like an F1 car where the matching process is variable and controllable (The driver can choose the amount of power the electric motor provides and change it so that he has more or less available on different parts of the track).
     
  16. Apr 27, 2017 #15
    Think of trying to move a large boulder by using a lever. Apply 700 ft-lb to the boulder through the lever. Lets say it does not move. Now take another lever arranged identically, but right next to the first lever. Apply 400 ft-lb through that lever to the boulder, while 700 ft-lb is still being applied to the boulder through the first lever. You are currently applying 1100 ft-lb to the boulder.

    If it still doesn't move, you're going to need a smaller boulder.
     
  17. Apr 29, 2017 #16
    I can imagine now thank you all for answers ! :)
     
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