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Tractive effort

  • 09-03-2004 7:57pm
    #1
    Closed Accounts Posts: 439 ✭✭


    In how it applies to trains and the like, I just don't understand what it is, or how to calculate it, can't seem to find a book that even defines it, but we're expected to work it out. The example we have pulls numbers out of no where, If someone could explain the principals that would be great.


Comments

  • Registered Users, Registered Users 2 Posts: 33,518 ✭✭✭✭dudara


    I was curious about this when I saw the post, so I had a look around. I had guessed it was something to do with pulling ability (no puns intended) of the train. I hope these help. It seems to be essentially related to the power that a locomotive can deliver to the rails through it's wheels, slippage aside.

    Link 1

    Link 2

    Link 3


  • Closed Accounts Posts: 439 ✭✭Atreides


    A steam locomotive’s boiler delivers a pressure of 15 bar to 2 double acting cylinders, bore 490 mm stroke 980 mm, which act on driving wheels 2 m in diameter. Calculate the theoretical maximum tractive effort (in tonnes) with regulator and cut-off fully open.

    Each time the driving wheel rotates, it travels 2 pie metres, so the work done is 2 x pie x 9810 Joules if T is in tonnes.

    At the same time, each piston travels a distance of 2x0.98 metres, under a force of 15x105xpx0.492/4 Newtons.

    therefore tracktive effort equals 18 tones. Now I understand where he got 18 from (the maths involved) but i don't understand where he got his figures for work don't force or distance travelled. Could someone be good enough to explain to me where they came from.


  • Registered Users, Registered Users 2 Posts: 33,518 ✭✭✭✭dudara


    15x105xpx0.492/4 Newtons

    Ok, I'm not sure, but I think that you are calculating the force in the cylinder here.

    Maybe it should read (15e5)(PI*r^2) or (Pressure)(Area). The bore is 0.49m, therefore, the radius needed is 0.49/2. Is this any use?


  • Closed Accounts Posts: 439 ✭✭Atreides


    sorry yes that should be 15x(10^5) x pie x(0.49^2)/4 Newtons.

    I get the 15x(10^5), thats converting into pascals.

    thats exactly it, thank you so much i didn't realise it was presure by area, as I said he didn't explain to well. Where did the work one come from now, its mulitplied by two due to two cylindres.

    and the distance is .98meters by 2 because its double acting


  • Registered Users, Registered Users 2 Posts: 33,518 ✭✭✭✭dudara


    well work W = (Force)(Distance), so the work done by the cylinders is 2 (Force)(Distance travelled)


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  • Moderators, Recreation & Hobbies Moderators, Science, Health & Environment Moderators, Technology & Internet Moderators Posts: 93,596 Mod ✭✭✭✭Capt'n Midnight


    Oddly enough in tractor pulling contests (the sled thing) most of the power does not come from tractive effort - it actually comes from the "recoil" of the dirt being sent backwards by the tires (conservation of momentum and all that.)

    Bit like the SR71 Blackbird - only 18% of the thrust comes from the engines !


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