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Just need a "simple" question answered around something I'm designing it's mechanical in nature but confident there are people who may be able to help

Maglatron

Jul 12, 2023
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Gears, flywheels, levers, falling blocks, gravity, inertia, inertia force, momentum, angular momentum, velocity, angular velocity, acceleration, angular acceleration, displacement, energy, force, torque, power, mechanical advantage, all the good stuff, although not sure this is the place for it I'm giving it a try here because the guys here are ace

It's mechanical in nature


please follow, I will write down the steps one by one.





  1. Starting with the flywheel on the left side of the drawing, the flywheel has a gear on the axle of it.
  2. The lever with load side length of 0.005m is moved through 20 degree (0.3940658504 rad)

    and moves counter clockwise (the flywheel moves in the clockwise direction)




2) the arc length of load side = 0.001745329252m





3) inertia = 10.8 because weight is 60kg and radius is 0.6m





4) then we actually make the arc length that the load side of lever makes the circumference of the gear that attaches to axle of the wheel! so its radius is the arc length over 2pi which equals 0.0002777777778m





5) tangential force on gear equals torque over radius so pick arbitrary acceleration 0.05 multiply by inertia 10.8 and the torque equals 0.54. then 0.54/0.0002777777778 equals force of 1944 newtons





point 5 continued - torque also equals F * r so; 1944 * length of the load side of lever (0.005) multiply and arrive at 9.72Nm of torque. divide this number by an arbitrary length of 0.3m so, 9.72/0.3 gives the force in newtons to achieve the force that needs to act on the end of the effort side of the lever which is 32.4N





6) find the distance that the effort side of lever make, so (s = r * rad) 0.3 * 0.3490658504 rad = s of 0.1047197551m, and actually make this length be the circumference of a driving gear on a compound gear (the smaller one is the driven gear) the radius of the driving gear is the circumference/2pi which turns out to be 0.1047197551/2pi = 0.01666666667m.





7)for clarification multiply force by radius of driving gear 32.4 by 0.01666666667 that equals 0.54Nm





8) make driven gear on compound 5 * smaller by dividing the radius by 5 0.01666666667/5 = 0.003333333334, and then force 5 * larger: 32.4 * 5 = 162N





9) multiply 162 * 0.003333333334 = 0.54





10) make second gear on compound (smaller gear) find its circumference 0.003333333334 * 2pi = 0.02094395103m and make lever two load side move this distance s = r * rad so s/rad equal radius or distance to pivot 0.02094495103/0.4390658504 = 0.06m





11) times the 0.06 by 60 because that was the original ratio of the first lever so keeping it all neat = 3.6m





12) next to find the force needed on the end of lever 2, divide the newtons on the load side by 60 because that's the ratio 162/60 = 2.7N





13) divide 2.7 by the gravitational constant 2.7/9.80665 = 0.2753233775kg





14) find height that the effort side of the 2nd lever when raised through 0.3490658504rad, so 3.6 * 0.3490658504rad = 1.256637061m





15) torque equals F *r. 2.7 * 3.6 = 9.72Nm as calculated earlier!!





16) energy; F * d 2.7 * 1.256637061 = 3.392920065Joules torque time theta 9.72 * 0.3490658504 = 3.392920066J mgh 0.2753233775 * 9.80665 * 1.256637061 = 3.392920065J to work out the equation for E = 1/2 I * w^2 we must find the angular velocity w2^2 = w1^2 + 2 alpha * theta (w1 is 0) = 2 * 0.05 * 2pi rad = 0.7926654595 take the root and it gives 0.8903176172rad/s so plug into energy equation E = 1/2 * 10.8 * 0.7926654595^2 which indeed equals 3.392920099J





  1. Initial Setup:
    • A lever with a load side length of 0.005 meters is moved to an angle of 20 degrees (0.3940658504 radians).
    • This results in an arc length of 0.001745329252 meters.
  2. Inertia Calculation:
    • The inertia is determined to be 10.8 based on a weight of 60 kg and a radius of 0.6 meters.
  3. Gear and Torque Calculations:
    • The arc length of the lever is used to create the circumference of a gear attached to the wheel axle.
    • The torque required to move the gear is calculated based on an arbitrary acceleration, resulting in a force of 1944 newtons.
    • Torque is also calculated using the force exerted on the lever side, resulting in 9.72 Nm.
    • The force needed on the effort side of the lever is calculated to be 32.4 N.
  4. Compound Gear System:
    • The height of the effort side of the lever is found to be 0.1047197551 meters, which is used as the circumference of a driving gear.
    • The radius of the driving gear is calculated to be 0.01666666667 meters.
    • The driven gear is made smaller by dividing its radius by 5, resulting in a radius of 0.003333333334 meters.
    • The force on the driven gear is calculated to be 162 N, and the torque is confirmed to be 0.54 Nm.
  5. Second Compound Gear System:
    • The circumference of the smaller gear in the compound system is calculated.
    • The distance to the pivot point of the lever is determined based on the gear circumference.
    • The force needed on the end of the second lever is calculated to be 2.7 N.
  6. Energy and Angular Velocity Calculations:
    • Energy is calculated using different methods: force times distance, torque times angle, and potential energy.
    • The equation for rotational kinetic energy, E = ½ * I *ω^2, is derived and confirmed.

    • ok good but at this point I want to point out that the end of the of the second lever is raised first and when it drops it raises the second lever so the wheel would spin clockwise

    • Equation: Theta (in radians) = Initial angular velocity (w1) * time (t) + 1/2 * Angular acceleration * time^2


      Given values: Initial angular velocity (w1) = 0 Angular acceleration = 0.05 Theta (angle for one full revolution) = 2pi


      Solution: 2pi = 0 * t + 1/2 * 0.05 * t^2


      0.05 * t^2 = 2pi * 2


      t^2 = (2pi * 2) / 0.05


      t^2 = (4pi) / 0.05


      t^2 = (4 * 3.14159) / 0.05


      t^2 ≈ 12.56636 / 0.05


      t^2 ≈ 251.3272


      t ≈ sqrt(251.3272)


      t ≈ 15.85330919 seconds

    • Here's a summary of the drops along with the corresponding time durations, angular velocities, and rotational kinetic energies:


      Drop 1:
      • Time: 15.85330919 sec
      • Angular velocity: 0.8903176172 rad/sec
      • Rotational kinetic energy: 3.392920066 J
      Drop 2:
      • Time: 6.034654264 sec
      • Angular velocity: 1.1920533 rad/sec
      • Rotational kinetic energy: 7.673351778 J
      Drop 3:
      • Time: 4.789753763 sec
      • Angular velocity: 1.431540988 rad/sec
      • Rotational kinetic energy: 11.06627184 J
      Drop 4:
      • Time: 4.096099625 sec
      • Angular velocity: 1.636345969 rad/sec
      • Rotational kinetic energy: 14.45919192 J
      Drop 5:
      • Time: 3.637605426 sec
      • Angular velocity: 1.81822624 rad/sec
      • Rotational kinetic energy: 17.85211197 J
      Drop 6:
      • Time: 3.305439609 sec
      • Angular velocity: 1.98349822 rad/sec
      • Rotational kinetic energy: 21.24503203 J
      Drop 7:
      • Time: 3.05044627 sec
      • Angular velocity: 2.136020534 rad/sec
      • Rotational kinetic energy: 24.63795209 J
      Drop 8:
      • Time: 2.846692596 sec
      • Angular velocity: 2.278355164 rad/sec
      • Rotational kinetic energy: 28.03087216 J
      Drop 9:
      • Time: 2.679018985 sec
      • Angular velocity: 2.412306113 rad/sec
      • Rotational kinetic energy: 31.42379223 J
      Drop 10:
      • Time: 2.537888269 sec
      • Angular velocity: 2.539200526 rad/sec
      • Rotational kinetic energy: 34.81671229 J
      Drop 11:
      • Time: 2.1416958802 sec
      • Angular velocity: 2.660048466 rad/sec
      • Rotational kinetic energy: 38.20963235 J
      Drop 12:
      • Time: 2.311826942 sec
      • Angular velocity: 2.775639813 rad/sec
      • Rotational kinetic energy: 41.60255241 J
      Drop 13:
      • Time: 2.219326175 sec
      • Angular velocity: 2.886606122 rad/sec
      • Rotational kinetic energy: 44.99547247 J
      Drop 14:
      • Time: 2.13711318 sec
      • Angular velocity: 2.993461718 rad/sec
      • Rotational kinetic energy: 48.38839255 J
      Drop 15:
      • Time: 2.063411596 sec
      • Angular velocity: 3.096632298 rad/sec
      • Rotational kinetic energy: 51.78131057 J
      Drop 16:
      • Time: 1.996846826 sec
      • Angular velocity: 3.196474639 rad/sec
      • Rotational kinetic energy: 55.17423064 J
      Drop 17:
      • Time: 1.936336444 sec
      • Angular velocity: 3.293291461 rad/sec
      • Rotational kinetic energy: 58.5671507 J
      Drop 18:
      • Time: 1.881014774 sec
      • Angular velocity: 3.3873422 rad/sec
      • Rotational kinetic energy: 61.96007076 J
      This data provides a detailed look at how the angular velocity and rotational kinetic energy of the wheel change over successive drops.

so now for the question!!!
I want to change gear so that the flywheel remains accelerating at 0.05rad/sec^2 but so that the block takes 15 seconds to drop on the 19th time the block is dropped. the flywheel is free to freewheel and the block is manually lifted back to the top to it's starting posistion each time before it's dropped again. Thanks people!
 
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Maglatron

Jul 12, 2023
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yeah also forgot that there is an idler gear between the 2nd lever and the smaller driven gear on the compound gear to make the flywheel rotate clockwise drew out the diagram and realized also it's massively out of scale but I'll get the camera tomorrow maybe and post the drawing, thanks might sit for a while and draw it in paint!
 

Maglatron

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so I make a diagram which I think is sufficient, oh yeah also the blue arc that touches the yellow, the length of that is the circumference of the smaller gear on the compound (orange) so when the weight moves the 2nd lever through an angle of 20 degree the compound gear rotates once! there used to be a function in paint where a little green dot appeared and you could rotate does it not have that anymore it would be good for labling hope you can understand whats going on with it! thanks
 

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Maglatron

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Wiki will have reference to all your queries I am sure.
thanks but can someone have a crack at it because finding it difficult to find the right info, unless you can point me to the correct page?!? thanks
 

Maglatron

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an analogy would be a car accelerating in 1st gear from 0-20mph in 5 second in 1st from 2000rpm to 3000rpm, then change to 2nd gear and accelerate from 20-30mph in 5 seconds in 2nd gear from 2000rpm to 3000rpm, then change to 3rd gear and accelerate from 30-40mph in 5 seconds in gear 3 from 2000 to 3000rpm
the engine slows down when the gear is changed! hope you get what I'm saying!! thanks
 

bertus

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Hello,

You might find some info on the following site :


Bertus
 

Maglatron

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found the gear section going to take a deeper look, thanks
 

Maglatron

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does any one know what I'm even on about by "keeping the acceleration the same but making the block take longer to fall" I appreciate the comments
 

Maglatron

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before I can understand I need to know the theory behind how it works! I want to keep the angular acceleration 0.05rad/sec^2 whilst extending the time it takes for the block to fall
 

Harald Kapp

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@Maglatron : Please stop sending each sentence in a separate post. It is not difficult at all to combine multiple sentences in a single post.
 

Maglatron

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I'm having a great deal of difficulty finding info on changing gears and I can't find a single example of what I have here I appreciate the engineering toolbox site but don't know how or even if I can apply the information found over there with what I'm doing. preferably I'd like someone who is an expert in the field of mechanics/dynamics!

I have a table and I also want to add into it the velocity just before the block hits the floor on each succesive drop of the weight, thanks
and I want to be able to change gear between the block - lever2 - compound gear - lever1 - flywheel. as to have the weight take longer to fall whilst still having the flywheel accelerate at 0.05rad/sec^2 hope you have enough infor and if you have question I'm here to answer as best I can!


so Iv'e written it out again I want my flywheel whose inertia is 10.8kgm^2 and initial velocity is 3.3873422rad/sec it has a gear on the axle of it whose radius is 0.0002777777778m lever, whose load side is 0.005m and effort side 0.3m (this is to the left of the wheel) and the arc length on the load side is 0.001745329252m because the and it moves through is 0.3490658504rad * the length of the load side lever 0.005 and 0.001745329252m is the circumference of the gear on the axle of the flywheel the effort side moves up which in turn moves the wheel clockwise. an arbitrary angular acceleration is chosen as to be 0.05rad/sec^2. torque equation states inertia * angular acceleration equals torque 10.8 * 0.05 = 0.54Nm. torque/radius of gear equals tangential force 0.54/0.0002777777778 = 1944N. then 1944 * length of load side of the lever = torque so 1944 * 0.005 = 9.72Nm then to find the force needed on the end of the lever (over to the left more) 9.72/0.3 = 32.4N moving upwards, find length of arc on effort side of lever one, s = r theta 0.3 * 0.3490658504 = 0.1047197551m, then we make that distance the circumference of the driver gear on a compound gear the bigger on is the driver and the smaller one is the driven so 0.1047197551m/2pi equals the radius of the larger gear on the compound which is 0.01666666667m. make the driven gear 5 times smaller and multiply the force by 5, so 32.4 * 5 = 162N and the radius of the smaller gear is 0.01666666667/5 = 0.003333333334m, then for continuity multiply 162 * 0.003333333334 = 0.54Nm add idler gear between the smaller gear on the compound and the toothed portion of the arc of the load side on lever two to make the flywheel turn clockwise th size of this gear is unimportant because it's an idler gear and make the circumference of the driven gear the the distance the load side make 0.003333333334 * 2pi = 0.02094395103 s = r theta, s / theta = r , 0.02094395103/0.3490658504 = 0.06m. multiply by 60 because that was the ratio of the first lever and that quals 3.6m for the 2nd lever effort side. next find the distance of that the end of the ffort side is lifted s = r theta, 3.6 * 0.340658504 = 1.256637061m. then divide 162/60 = 2.7N so the weight of the block that pulls down on the end of the second lever 2.7/9.80665 = 0.2753233775kg then we have torque match up with 2.7N * 3.6m = 9.72Nm, then for energy; F * d 2.7 * 1.256637061 = 3.392920065Joules torque time theta 9.72 * 0.3490658504 = 3.392920066J mgh 0.2753233775 * 9.80665 * 1.256637061 = 3.392920065J to work out the equation for E = 1/2 I * w^2 we must find the angular velocity w2^2 = w1^2 + 2 alpha * theta (w1 is 3.293291461) = 3.293291461^2 + 2 * 0.05 * 2pi rad = 11.47408718 take the root and it gives 3.3873422rad/s so plug into energy equation E = 1/2 * 10.8 * 3.3873422^2 which equals 61.96007077J analyse and summarize

the above skips the first 18 drops and so starts with the angular velocity at 3.3873422rad/s

I would like to know the velcities of the block momentarily before it hits the ground and add to the table below and I want to know how to change gear on the 19th drop of the weight to maintain the angular acceleration 0.05rad/sec^2 but I want the weight to take 15sec to drop to the ground when it has done to 20 degree (0.3490658504rad)

Drop 1:
  • Time: 15.85330919 sec
  • Angular velocity: 0.8903176172 rad/sec
  • Rotational kinetic energy: 3.392920066 J

Drop 2:
  • Time: 6.034654264 sec
  • Angular velocity: 1.1920533 rad/sec
  • Rotational kinetic energy: 7.673351778 J

Drop 3:
  • Time: 4.789753763 sec
  • Angular velocity: 1.431540988 rad/sec
  • Rotational kinetic energy: 11.06627184 J

Drop 4:
  • Time: 4.096099625 sec
  • Angular velocity: 1.636345969 rad/sec
  • Rotational kinetic energy: 14.45919192 J

Drop 5:
  • Time: 3.637605426 sec
  • Angular velocity: 1.81822624 rad/sec
  • Rotational kinetic energy: 17.85211197 J

Drop 6:
  • Time: 3.305439609 sec
  • Angular velocity: 1.98349822 rad/sec
  • Rotational kinetic energy: 21.24503203 J

Drop 7:
  • Time: 3.05044627 sec
  • Angular velocity: 2.136020534 rad/sec
  • Rotational kinetic energy: 24.63795209 J

Drop 8:
  • Time: 2.846692596 sec
  • Angular velocity: 2.278355164 rad/sec
  • Rotational kinetic energy: 28.03087216 J

Drop 9:
  • Time: 2.679018985 sec
  • Angular velocity: 2.412306113 rad/sec
  • Rotational kinetic energy: 31.42379223 J

Drop 10:
  • Time: 2.537888269 sec
  • Angular velocity: 2.539200526 rad/sec
  • Rotational kinetic energy: 34.81671229 J

Drop 11:
  • Time: 2.1416958802 sec
  • Angular velocity: 2.660048466 rad/sec
  • Rotational kinetic energy: 38.20963235 J

Drop 12:

  • Time: 2.311826942 sec
  • Angular velocity: 2.775639813 rad/sec
  • Rotational kinetic energy: 41.60255241 J

Drop 13:
  • Time: 2.219326175 sec
  • Angular velocity: 2.886606122 rad/sec
  • Rotational kinetic energy: 44.99547247 J

Drop 14:
  • Time: 2.13711318 sec
  • Angular velocity: 2.993461718 rad/sec
  • Rotational kinetic energy: 48.38839255 J

Drop 15:
  • Time: 2.063411596 sec
  • Angular velocity: 3.096632298 rad/sec
  • Rotational kinetic energy: 51.78131057 J

Drop 16:
  • Time: 1.996846826 sec
  • Angular velocity: 3.196474639 rad/sec
  • Rotational kinetic energy: 55.17423064 J

Drop 17:
  • Time: 1.936336444 sec
  • Angular velocity: 3.293291461 rad/sec
  • Rotational kinetic energy: 58.5671507 J

Drop 18:
  • Time: 1.881014774 sec
  • Angular velocity: 3.3873422 rad/sec
  • Rotational kinetic energy: 61.96007076 J

I'm having a great deal of difficulty finding info on changing gears and I can't find a single example of what I have here I appreciate the engineering toolbox site but don't know how or even if I can apply the information found over there
 

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Maglatron

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its like peddling like mad in first on a push bike and then change gear so that its a nice slow motion smooth motion, thanks
 

Maglatron

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and it's also like you can't change gear from 1 to 6, you have to go in succession through the gears so the gear has to match the speed, if you catch my drift
 

Delta Prime

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I'm very fond of this site!
I'm not a mechanical engineer but I played one on TV.:p
 

Maglatron

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thanks Mr Delta Prime but I'm banned from that site because I was asking about my inverter way back when I was on that project and they were saying that and something about not being able to do it and they wouldn't like to be part of eliminating myself from the gene pool using voltages that in their opinion, was not qualified to be working with high voltages can't remember the rest of why I was banned but they won't let me back on!!!
just tried to log in on there and it says You have been banned for the following reason: Sockpuppet of banned member checkmatescott. also rules violation
which I thought was really crappy because I was a legitimate user
 
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Delta Prime

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Could you provide a link to the photo you posted about (project jpeg).
I'm having difficulty following your simple question thank you.
 

Alec_t

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Please be sensible about numbers used. It is totally impractical to make a mechanism as you describe to an accuracy quoted to nine decimal places!
 
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