Engineering/physics formula questions about a Supersplit

   / Engineering/physics formula questions about a Supersplit #21  
LD1, is the flywheel actually a disk or is the flywheel weighted to the outer edge? (skinny hub or spokes with heavy outer ring). If so then you might have to use the formula for a thick hoop I = 1/2M(a^2 + b^2) where a and b are the inner and outer radius of the rim portion; Then add the disk formula for the hub / spoke part. You would have to guestimate the mass of each (hub / rim).
 
   / Engineering/physics formula questions about a Supersplit
  • Thread Starter
#22  
LD1, is the flywheel actually a disk or is the flywheel weighted to the outer edge? (skinny hub or spokes with heavy outer ring). If so then you might have to use the formula for a thick hoop I = 1/2M(a^2 + b^2) where a and b are the inner and outer radius of the rim portion; Then add the disk formula for the hub / spoke part. You would have to guestimate the mass of each (hub / rim).

They are more disk like. The SS flywheels do have some lightning holes in them. But the homemade ones that I have seen, are solid disks.
 
   / Engineering/physics formula questions about a Supersplit #24  
The spur and rack have to have the same diametral pitch. So it dont matter weather they are 6 pitch, or 32 pitch. A 3" spur gear is going to move its mating rack that same 9.4 inches.

Now if we were talking a 18tooth spur of a different pitch, that would be a different story. But the diameter would NOT be 3" and therefore the circumfrence would NOT be 9.4.
I responded in reference to the "ratio" being 1:1 - does'nt the size gear you have create an actual ratio of 1 rev. equals 9.4" linear movement? In other words, would not a larger gear that does not require a full revolution to deliver 9.4" of movement be yet a different ratio? Also requiring less effort to move the ram 9.4" ?
 
   / Engineering/physics formula questions about a Supersplit #25  
I responded in reference to the "ratio" being 1:1 - does'nt the size gear you have create an actual ratio of 1 rev. equals 9.4" linear movement? In other words, would not a larger gear that does not require a full revolution to deliver 9.4" of movement be yet a different ratio? Also requiring less effort to move the ram 9.4" ?

A larger gear would move the gear the same distance using less rotation. Mechanical advantage (of the flywheel in this case) would be reduced when using a larger gear.

A smaller gear would require less effort, rack would move slower and have more force.
 
   / Engineering/physics formula questions about a Supersplit #26  
A larger gear would move the gear the same distance using less rotation. Mechanical advantage (of the flywheel in this case) would be reduced when using a larger gear.

A smaller gear would require less effort, rack would move slower and have more force.
Aha, my thinking is backwards LOL!
 
   / Engineering/physics formula questions about a Supersplit #27  
I do not think a 3 inch gear will translate in to the distance of the out side diameter, but what would be about at the diameter of the center of the tooth, because it pushes at the tooths center not it tip unless it is on the entering or exiting, teeth of the pinion, the diameter would be the pitch circle, File:Gear words.png - Wikipedia, the free encyclopedia it has been a long time since I have studied gears, but

as the gear mesh there is a over lap that is do to the teeth of the gear, and there is the pitch circle which is the rolling diameter of the gear,

(I really doubt if this has any bearing on the discussion at hand or actual function of the splitter, but there is discussion on how far the pinion would move on a 3" gear, and it would depend on the pitch circle not the gears over all diameter, a finer toothed gear that was 3" over all, would move it faster than a coarse tooth gear that had big deep teeth as they both would have a different pitch circle, (in this instance the large coarse tooth would most likely be more robust and would deal with the shock load better, and would handle dirt and dust better),
 
   / Engineering/physics formula questions about a Supersplit
  • Thread Starter
#28  
I do not think a 3 inch gear will translate in to the distance of the out side diameter, but what would be about at the diameter of the center of the tooth, because it pushes at the tooths center not it tip unless it is on the entering or exiting, teeth of the pinion, the diameter would be the pitch circle, File:Gear words.png - Wikipedia, the free encyclopedia it has been a long time since I have studied gears, but

as the gear mesh there is a over lap that is do to the teeth of the gear, and there is the pitch circle which is the rolling diameter of the gear,

(I really doubt if this has any bearing on the discussion at hand or actual function of the splitter, but there is discussion on how far the pinion would move on a 3" gear, and it would depend on the pitch circle not the gears over all diameter, a finer toothed gear that was 3" over all, would move it faster than a coarse tooth gear that had big deep teeth as they both would have a different pitch circle, (in this instance the large coarse tooth would most likely be more robust and would deal with the shock load better, and would handle dirt and dust better),

You are correct in the term pitch diameter. But you are right that it doesn't really apply to what I am talking about. Because with a 6 pitch 18tooth gear, I refered to it as a 3" diameter gear, but actually that 3" IS the pitch diameter. That is how the pitch or 6 is determined. The # of teeth divided by the Pitch diameter = the pirch if the gear. SO....18teeth/3" pitch diameter = a 6 pitch gear.

I am saying is that if you take a 36 tooth 12 pitch gear (36/12=3) which also has a pitch diameter of 3", 1 full revolution will move the corrosponding rack the exact same distance as the 18tooth 6pitch gear would.

Clear as mud right:laughing:
 
   / Engineering/physics formula questions about a Supersplit #29  
A larger gear would move the gear the same distance using less rotation. Mechanical advantage (of the flywheel in this case) would be reduced when using a larger gear.

A smaller gear would require less effort, rack would move slower and have more force.
The same force would be available and act over the same ram travel distance. With the bigger gear the flywheel would just not rotate as far while stopping - ie, it would decelerate more quickly while delivering its energy.
larry
 
 
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