+-

Greeting

Welcome to my simple forum
 
Please be considerate of all members
Cookies and Java-Script are not needed
but can be used for YOUR convenience
I do not have ads on this site so do not place any on it
I have allowed registration upon my approval
the solution is
one is 1
 

User

Welcome, Guest.
Please login or register.
 
 
 
Forgot your password?

+-Stats ezBlock

Members
Total Members: 6
Latest: gdad
New This Month: 0
New This Week: 0
New Today: 0
Stats
Total Posts: 523
Total Topics: 47
Most Online Today: 95
Most Online Ever: 1092
(July 26, 2026, 11:26:43 pm)
Users Online
Members: 0
Guests: 92
Total: 92

Author Topic: Maxwell's Engine  (Read 3942 times)

webby2

  • Hero Member
  • *****
  • Posts: 521
    • View Profile
Re: Maxwell's Engine
« Reply #15 on: August 13, 2026, 04:18:28 am »
The printer I used for the frame sections had an issue that I have now resolved. I reprinted those parts and also made the dual ring gear, or hybrid ring gear, a little stronger so it is less prone to deforming under load. I also found two bad pivot bearings and replaced them.

As tested:

28 bearings
17 gears

Blue motor alone: 0.07 A

1:1 direct head-to-head, 1 Blue driving 3 Yellow:
0.48–0.50 A

2:1 system, no load motors:
0.16–0.23 A

1:1 system, no load motors:
0.18–0.24 A

2:1 head-to-head through the system, 1 Blue driving 3 Yellow:
0.45–0.54 A

All tests were run at the same 6 V power-supply setting and allowed to run until reasonably stable.

These are the measured numbers from this build. They are not being presented as evidence of gain; they are simply the results from the current configuration.

webby2

  • Hero Member
  • *****
  • Posts: 521
    • View Profile
Re: Maxwell's Engine
« Reply #16 on: Today at 01:29:59 am »
something that has been bother me with this is that the math only shows 1/2 of the input coming back to assist and I did not see why that would be, well I finally figured that out and it is so silly that it is embarrassing that I missed it.

There are 3 systems lets say,, there is the upper input section, the slip and then the lower return section.  The upper and lower share a physical connection on one side and a force connection on the other and it is that force only  connection where the slip is.

Now if the input goes in on the upper and it spins the upper and then the upper goes across the slip and spins the lower and then the lower comes back to assist the upper,, the simple thing is that the slip boundary has the upper traveling at say 100 RPM and with the 2:1 return gearing the lower is traveling at 50 RPM, so the same force but at half the rate means half the return work,,, and that is why there is only 1/2 of the input returned to assist.

 

Powered by EzPortal