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Thread: Newton, and newton meters

  1. #1 Newton, and newton meters 
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    the formula for a newton is 1 N = 1kg*m/s^2.
    but me, not being very experienced with physics, i can't understand how
    you can compress a rate into a single number?
    especially when calculating torque?
    i was trying to figure out the energy exerted onto a gun-tube on a tank(not taking into account the friction of the turret ring vs hull)

    the tank weights 62500kg and is speeding down the road at 65kph or 18.05m/s.
    the length from the tip of the gun tube to the center of the turret basket is around 6 meters.

    what i want to know, is if the turret would be able to rotate 360 degrees 15-20 times before coming to a stop.
    a friend of me says no. he says the force of impact equals to the amount of torque in the engine, which is 3754 Nm @ 3000rpm.

    i say maybe, but i don't have much else than the second-hand accounts of some crewmen delivered to a guy who wrote a book.
    but the centrifugal force was high enough to force an 80kg man into the turret wall for several seconds in what he described as "a gravity ride from a carnival."

    the turret is on greased ball-bearings to minimise friction, so i guess the friction number is around 0.01 to 0.1
    the turret drive is hydraulic, but exploded due to the impact, so the hydraulic pressure of the system equals 0, allowing the turret to rotate freely. the only rotational restriction is the friction.

    the formula i have for the torque is currently

    X Nm = 6m * (62500kg*18.05m/s^2)

    how do i solve this? do i just calculate the numbers as usual?

    is X = 6*(62500*18.05/1^2) ?
    how do i make a graph, of the initial speed, and its decrease based on the friction number?
    its a bit confusing working with a number thats essentially a rate, that increases per second.


    when you have eliminated the impossible, whatever remains, however improbable, must be the truth
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  3. #2  
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    I don't get it. What is impacting what and why? What is making the turret spin? Does the barrel run into something? Are you calculating an energy or a torque?


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  4. #3  
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    you need to know WHY its impacting? why do you need to know why its impacting? it was an accident.
    oh i forgot to write what it was impacting. a concrete block from a highway overpass.
    still, your post is remarkably lacking in explanations.
    i wrote several questions, both pertaining the final results, and how to calculate the formula for the newton.
    when you have eliminated the impossible, whatever remains, however improbable, must be the truth
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  5. #4  
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    scenario looked something like this.



    according to an account, the turret span around 15-20 times,
    and the vehicle kept moving. the purpose is to find out if
    its possible for a tank turret, based on the impact, to spin 15-20 times.
    now, any infomration you need, i can provide.
    when you have eliminated the impossible, whatever remains, however improbable, must be the truth
    A.C Doyle
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  6. #5  
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    is:

    6m * (62500kg*18.05m/s^2) = 6*(62500*18.05/1^2)
    when you have eliminated the impossible, whatever remains, however improbable, must be the truth
    A.C Doyle
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  7. #6  
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    Would it it not be better to work with momentum formulas? Break V into vector components and work out an initial velocity that the turret would spin at, and then use a simple accelleration formula to work out how long it would take for friction to stop it. I guess the hydrolic system would eat some momentum before exploding and maybe a short instant of friction between the barrel and the concrete block before the elastic rebound pushed it away. If I'm talking shite, please forgive my ignorence.

    Edit: guess you'd need the weight of the turret for that.
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  8. #7  
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    yes, the hydraulic system did last enough cause a gash in the concrete, but
    it gave in, and the turret started spinning wildly.

    i'm new to trying to using formulas for proving something, and have very little in the way of mathematical usage of formulas, sometimes i'm not certain what to replace in the formulas i find on wikipedia.
    like in this case. and i'm completely clueless on friction calculations,
    and how torque mixes with those.

    i tried to find out the speed of a vehicle where X torque was being put to a wheel, and the ground-wheel had a certain friction, but couldn't link these.
    when you have eliminated the impossible, whatever remains, however improbable, must be the truth
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    It looks like it’s up to me to answer this one. I guess all the other physics whizbangs are too busy solving the mysteries of black holes and the big bang. That’s okay. I like this kind of problem.

    As you have discovered, you can’t just find a formula and start plugging and chugging. You have to understand some of the physics behind it.

    Your friend is barking up the wrong tree by trying to calculate the engine torque, thrust on the tank tracks or anything like that. You might do that if you have the tank parked with its gun barrel against the bridge pier and try to start moving. A collision is a whole different thing. To see why this is true think about what happens when you hit a nail with a hammer. The torque or force you can apply to the hammer with your wrist is probably only a few foot-pounds or pounds of force on the hammer head. Certainly not enough to drive a nail. But when you hit the nail with the hammer the momentum of the hammer head changes almost instantaneously. That means the force gets very high. The actual forces during the collision are pretty hard to calculate, so we usually try to solve collision problems by looking at the before and after conditions. You’ll see what I mean.

    You are also barking up the wrong tree because the mass of the tank is not going to be a factor, except for the fact that it is heavy enough that it will keep on going pretty much the same speed after the collision.

    The first thing about a collision is to decide if it is elastic or inelastic. Imagine your tank is parked next to the bridge pier and you swing the turret and smash the barrel into the pier. Will it bounce off at the same speed in the opposite direction (elastic) or does it come to a dead stop against the pier (inelastic). I’m going to guess it is mostly inelastic because it take out a chunk of concrete, so when the end of the barrel hits at 18 m/s it comes to a dead stop giving it a velocity of 18 m/s relative to the tank. If there is some elasticity it could be a little higher

    Now we calculate the rotational speed of the turret. A point 6 meters from the axis is going 18 meters per second. The circumference of a circle of radius 6 m is 2* pi*6 which is about 36 m so the turret is rotating at about half a revolution per second. By the time it comes to a stop if the deceleration is at a constant rate, the average speed of rotation will be ¼ revolution per second. So to make 20 revolutions it will come to a stop in 80 seconds. It doesn’t seem too far-fetched to me that the turret spins for a minute 20 seconds.

    How would we go about calculating it? We could use conservation of energy. The formula for rotational energy is
    E = 0.5 * I *(omega)^2
    Where I is the moment of inertia and omega is the angular speed in radians per second. The moment of inertia depends on the mass and the distribution of mass of the rotating body. You can look up the moment of inertia for different shapes. The tank turret will be rather complex consisting of a cylindrical rod (the barrel) on one side and the counterweight on the other side.

    Then we could equate the rotational energy to the energy lost to friction where the work of friction is equal to the frictional force multiplied by the distance through which that force is applied. For that we need to know the weight of the turret, the position of the turret bearing, and the coefficient of friction (which you have estimated as 0.1).

    If you can fill in any of that data, I will continue this tomorrow. Otherwise I could just make some educated guesses.
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  10. #9  
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    the weight of the turret is about 21 500kg, and the bearing is around 3 o'clock, or 90 degrees to the right. oh and thanks!
    when you have eliminated the impossible, whatever remains, however improbable, must be the truth
    A.C Doyle
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    I think your estimate of the coefficient of friction is way too high. Rolling friction is generally on the order of 0.001 not 0.1, but I will use your number anyway. If you decide it's too high, you can adjust the answer accordingly.

    The shape of the turret is fairly complex, and I don't have any dimensions so I will just assume its moment of inertia is given by a thin cylinder of mass m located at a radius of 2 meters. I don't need to use your estimate of the mass because it ends up canceling out in the final answer, since the frictional force is also proportional to m.

    I will assume that the bearing is located at a radius of 1 meter from the center of rotation. This radius is important because the further out from the center the more torque it exerts.

    So we have

    coeff of friction = 0.1
    r(inertia)=2 meters
    r(friction)= 1 meter
    r(barrel) = 6 meters
    v(barrel)=18 meters per second
    omega=v/r(barrel) = 3
    rotational energy E=0.5* I*(omega)^2
    I = m* r(inertia)^2 = 4*m

    E=0.5*4*m*3^2=18*m

    Work(friction) = m*g*coeff*X
    where X is the linear distance through which the friction is applied
    N= number of revolutions = X/(2*pi*r_friction)

    Equating the initial rotational energy to the energy lost to friction:
    18*m=m*9.8*0.1*X

    Then X=18/.98~18 meters which works out to 18/2pi=2.8 revolutions.

    But, as I said, I think your coefficient of friction is too high.
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  12. #11  
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    hmm, by bearing, do you mean the diameter/radius of the turret ring(contact point between hull and turret)?
    turret ring radius begins at about 108cm and ends around 128cm.

    the friction number is actually a range from 0.01-0.1
    i think the M1 turret is on ball bearings, which further reduces the friction.
    when you have eliminated the impossible, whatever remains, however improbable, must be the truth
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  13. #12  
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    i created a small program where i could calculate the friction based on this formula, written in C++

    #include <iostream>
    #include <math.h>

    using namespace std;

    int main()
    {
    float friction = 0.01; //friction number
    float Rinertia = 2.0; //meters
    float Rfriction = 1.0; //meters
    float Rbarrel = 6.0; //meters
    float Vbarrel = 18.0; // meters per second
    float omega;
    float X; //the linear distance through which friction is applied
    float I; //inertia i'd assume
    float E; //rotational energy
    float g = 9.8; //gravity it seems like.
    float N; //number of revolutions
    omega = Vbarrel/Rbarrel;
    I = pow(Rinertia,2);
    E = 0.5*I*pow(omega,2); //rotational energy
    X=E/(9.8*friction);
    N = X/(2*3.14*Rfriction);

    cout<< E << " rotational energy\n";
    cout<< X << " friction distance\n";
    cout<< N << " revolutions\n";
    system("pause");
    return 0;
    }

    based on this, if the friction is reduced to 0.01 the result is 29.2474 revolutions.
    with friction down at 0.001 you get 292.474 revolutions.
    when you have eliminated the impossible, whatever remains, however improbable, must be the truth
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  14. #13  
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    Quote Originally Posted by dejawolf
    hmm, by bearing, do you mean the diameter/radius of the turret ring(contact point between hull and turret)?
    turret ring radius begins at about 108cm and ends around 128cm.

    the friction number is actually a range from 0.01-0.1
    i think the M1 turret is on ball bearings, which further reduces the friction.
    Yes that's what I meant and it looks like my guess of 1 meter was close enough. We had to make some guesses about the mass distribution and the coefficient of friction, but as far as I'm concerned we did show that 15 to 20 rotations is feasible. The 292 revolutions seems a little bit too high. If the friction was that low, then the air resistance would start to become a bigger factor and would have to be considered.
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