Rack and Pinion Calculation help

[EngineeringDave]

Just to be aware this isn't homework this is a design concept and although calculations aren't required I'd like to assess a rough assessment of possible power generation figures.

The concept is a weight which pushes down upon the rack hence turning a pinion on a shaft which has lots of of these rack pinion systems, they are all connected to one shaft which in turn is connected to a flywheel/alternator system.

All material I've found to help myself, only looks at an motion input from the shaft not the rack and only looks to solve factor and safety issues (which of course are important).

Would anyone be so kind as to give some pointers or possible documents which may assist me.

Weight applied = 3300 N
Rack and Pinion dimensions we be reverse engineered from possible generator figures.

Regards

Dave

 

[OldEngr63]

Do I understand that you want a descending weight to drive the rack which in turn rotates the generator? What will you do when the rack gets to the end of its motion? What exactly is your question? A figure would probably help.

 

[tygerdawg]

Torque applied to shaft = (pitch diameter of pinion) X 3300N

 

[OldEngr63]

tygerdawg, are you sure you mean "pitch diameter"? Wouldn't pitch radius be more correct?

With the correct radius, this should give the torque applied to the shaft. It does not, however, address the separating force between the pinion and the rack.

 

[tygerdawg]

Yup, you're correct: pitch radius.

(It's the drugs. That's my excuse, and I'm sticking to it.)

 

[EngineeringDave]

I have drawn a figure here to assist.
Sorry about the grey scale.
Note: the box at the end would be an alternator or similar.

 

[billy_joule]

It's still not clear what exactly it is you'd like to know. If it's this:

I'd like to assess a rough assessment of possible power generation figures.

You can get an upper limit for power via
P = Fv

An upper limit for available energy can be found via:
E = mgh

Let's say your rack is 10 m tall
mg = F = 3300 N
Then
E = mgh = 3300 N * 10 m = 33,000 J = 33 kJ

Wolframalpha can give a sense of scale to that value:
~~ 1.2 × energy released by burning 1 gram of ethanol (~~ 27000 J )
~~ 0.84 × energy released by burning 1 gram of coal (~~ 39000 J )
~~ 0.87 × metabolic energy of one gram of fat (~~ 38000 J )

~~ 9.2 Wh (About enough to charge up one cellphone...)

http://www.wolframalpha.com/input/?i=33kJ