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Author Topic: Linear piston ratchet engine, 99% efficiency. Proto type in action!  (Read 29434 times)

Tommey Reed

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This is another engine design called the ratchet engine.
at 120 psi and a 12" stroke for each piston, this design converts linear into rotational energy.

At 12" stroke the formula for conversion is (24" x pi)/12)=6.28 ft and 120/12=10ft/lb for each revolution.
6.28 x 10 = 62.8 lb for each revolution of crank shaft for each 12cu/in of air.
At 600 rpm's this will produce (62.8 x 600)/33000= 1.14 hp
The amount of cfm's is (12x600)/1728 = 4.17 at 120 psi......


Proto-type in action.

http://www.youtube.com/user/OverUnityNow1#play/all/uploads-all/0/ccic7CH4sOY

Tom...

AlanA

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #1 on: September 03, 2009, 09:24:22 PM »
Hello Tommey,

I have seen you video and I have seen other videos about the ratchet engine which are no more available at youtube.
However I think that the ratchet engine is very powerful. In opposite to the Reciprocating engine the ratchen engine has no upper dead point and the lower dead point. The ratchet engine is powerful in every inch of its movement (ft lbs). Sadly the ratchet engine hasn't very much HPs. Is it possible to construct a larger engine with more HPs?

Could you please explain the formulas:
The stroke is 12 why is the figure 24" in the formule (24" x pi)/12). What means the 12 in the formula?
You have also written down this formula: (62.8 x 600)/33000. What means the 33000.
And also the formula for the cfms is not clear to me: (12x600)/1728. What means the 1728.

Please be patient with me. I am not so fit in maths als you.

Alana

Tommey Reed

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #2 on: September 03, 2009, 10:37:13 PM »
The Ratchet engine is calculated using many formulas.

We must first look at the piston size in cubic inches.

Lets start with say 2" diameter piston, the total is (3.14x(r^2))=3.14cu/in total piston area.

If we use a 4" gear on the output shaft, we can calculate how long the rack gear needs to be for 1 revolution per stroke:
the circumference of the 4" gear is (pi x d) or 3.14 x 4 =12.56in per stroke for each revolution of shaft.
 Now we have all the formulas to calculation torque:

At 100 psi on the piston with a area of 3.14cu/in we have 100 x 3.14=314 psi

having a 4" gear or 4"/2= 2" then we can take 314 psi x 2"= 628 psi or 628/12=52.33 foot pounds

At 12.56 piston stroke the will turn the shaft one full revolution at 6.28ft distant for each stroke the total output is 328.65 lb of work
At say 500 rpms we can calculate total work in one minute:
(500 x 328.65)/33000 =4.98hp
Or
(52.33ft/lb x 500rpm's)/5252=4.98hp

you can also calculate CFM's too:
3.14cu/in x 12.56" stroke =39.44 cu/in for each stroke
39.44cu/in x 500 rpm's =19719.2 cu/in in one minute

19719.2/1728=11.41 CFM's per minute that will produce 4.98hp at 100 psi of constant air pressure.....

I hope this helps you understand the true power of the Ratchet Engine...

Tom :-)



AlanA

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #3 on: September 04, 2009, 04:28:58 PM »
Thanks for the calculations. They clarify my questions.
Have you ever thought to reuse the air?
The ratchet engine is very powerful but isn't it a loss of energy to use the air only once?

Alana

Tommey Reed

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #4 on: September 04, 2009, 05:43:09 PM »
Any time you try to reuse air, its like puting a load on any system.
In other words, if I try to get 10 psi out of the air engine, I will also take 10psi from the intake pressure.
But, if I use lower pressure say 40 psi, then the compressors will be able to keep up with the CFM's.

Also if i use 300 psi at 1cu/in and inject it to to the engine, the expand air will still be able to finish with the piston moving 12.56 in. This will have 23 psi left over...

Tom...

AlanA

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #5 on: September 04, 2009, 07:17:16 PM »
Terry Miller could do this. He was able to reuse air in to, three and more cylinders. He could run several cylinders with the same air!!!

Alana

AlanA

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #6 on: September 12, 2009, 06:54:08 PM »
@ Tommey

Just to ensure that I understand you post right (Sorry Englih it not my first language).
I you use 300 psi at 1cu/in and inject it to to the engine, the air that come out of the cylinder is at 23 psi. I am right?

Thanks
Alana

Tommey Reed

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #7 on: September 12, 2009, 10:31:05 PM »
Just to ensure that I understand you post right (Sorry Englih it not my first language).
I you use 300 psi at 1cu/in and inject it to to the engine, the air that come out of the cylinder is at 23 psi. I am right?

Thanks
Alana

This all depends on many factors, If 300 psi  at 1 cu/in is in jected in to say 1cu/in piston each movment of the piston rod will travel at a psi rate of:
1" movement = 300 psi
2" movement = 150 psi
4" movement = 75 psi
8" movement = 37.5 psi
16" movement =18.75 psi
That is also if the air is turn off after 1cu/in of volume is used at 300psi injector...
I hope this helps...
Tom

Tommey Reed

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #8 on: September 12, 2009, 10:38:00 PM »
The 300psi injected in a 1cu/in piston will for the piston forward at a 12.56" stroke this is where the 23psi is from..
300psi/12.56= 23.8 psi....


Tom

AlanA

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #9 on: September 13, 2009, 07:07:19 PM »
Hi Tommey,

thanks for you brilliant answer.
Is there a formula to calculate the pressure loss?

Alana

P. S.: To come back to the calculation in your former posting:

I have not understand the line in your calculation above:

"having a 4" gear or 4"/2= 2" then we can take 314 psi x 2"= 628 psi or 628/12=52.33 foot pounds"

what means the /12 in the calculation?

Tommey Reed

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #10 on: September 13, 2009, 07:57:37 PM »
P. S.: To come back to the calculation in your former posting:

I have not understand the line in your calculation above:

"having a 4" gear or 4"/2= 2" then we can take 314 psi x 2"= 628 psi or 628/12=52.33 foot pounds"

what means the /12 in the calculation?




If we start with say 12 psi injected in to a piston area of 4", this piston have an area of (pi *(2^2))=12.56in/sq. 12psi x 12.56=150lb of force for each 1" of piston movement.

Lets say when building a ratchet engine, we start with a gear that is 4" diameter, gear has a circumference of (pi*4)=12.56in
So if say the rack moves 12.56 inches the gear will travel 360deg or 1 complete revolution.
The 6.28ft is the circumference of a 24" wheel to get a foot pound reading, if you have a constant of 10lb then you produced 62.8lb of work for each revolution of the shalf.

Tom

This is another prototype i'm working on.

http://www.youtube.com/user/OverUnityNow1#play/all/uploads-all/0/uANsNE2N8sk


AlanA

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #11 on: September 13, 2009, 10:08:29 PM »
Thanks for the answer and the link for your prototype of the ratchet engine. Very impressive.

In one of your former postings you have shown this chart:

This all depends on many factors, If 300 psi  at 1 cu/in is in jected in to say 1cu/in piston each movment of the piston rod will travel at a psi rate of:
1" movement = 300 psi
2" movement = 150 psi
4" movement = 75 psi
8" movement = 37.5 psi
16" movement =18.75 psi
That is also if the air is turn off after 1cu/in of volume is used at 300psi injector...

Is there a formula for the calculation of the pressure loss during the movement in the cylinder?

Alana

Tommey Reed

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #12 on: September 13, 2009, 11:43:52 PM »
We can look at a scuba tank to start with...
96cu/ft scuba tank at 3200 psi can be calculated on run time, it takes about 5hp compressor for about 20 minute to fill 96 cu/ft tank.
3200/14.7=217.7cu/ft if the scuba tank was 1cu/ft.
217.7/96=2.26 or 3200 scuba tank has an area of 3200/2.27=1409.7
1409.7/14.7=95.9cu/ft... 1cu/ft=1728cu/in....

1/2.27=.44 cu/ft of total area in a scuba tank, its not 1cu/ft at 3200psi.

So now we can calculated how long a engine can run on 1 scuba tank of 3200psi of air...

If we have a contant air flow of 12psi going into a piston say 1in/sq with 12" stroke for each revolution of the crank shaft.
If using my Ratchet Engine design this would produce 12ft/lb per revolution of shaft.
This will use (1" piston area x 12" stroke)=12cu/in
33000/12=2750 rpm's will make 1hp.
this will use a total of 33,000cu/in of air, or 33,000/1728=19.09cu/ft per minute
We know that a scuba tank has 1409.7 and now we can find out total storage at 12psi...
1409.7/12=117.475cu/ft total volume at 12psi.
117.475/19.09=6.15 minutes run time on 96cu/ft scuba tank..
The lost in not the engine,but the compressor charging the scuba tank.
At lower pressure like 12psi and small motor could make a constant feed of air to suppy this air engine...
If 1hp makes 4cfm at 90psi in one minute, then it should be able to make 19cfm's at 12psi constant.
Solar could be use to make 12 psi, even wind mills with a compressor could be charging a air tank...
Just some thoughts on how air can be made...

Tom








AlanA

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #13 on: September 17, 2009, 04:37:56 PM »
Thanks. You have answered more than I want to know.

12 PSI? Could this be real? This is a pressure less than the atmospheric pressure of about 14 PSI!

But I have another question you may could anser?
What means CUT OFF in the context of air engines?

Alana

Tommey Reed

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Re: Linear piston ratchet engine, 99% efficiency. Proto type in action!
« Reply #14 on: September 17, 2009, 08:32:02 PM »
Cut Off, could mean that you inject a volume of compress air, say 1cu/in at 200 psi. This could then be allowed to expand  until all that energy is zero.... So 200 psi at 1cu/in volume wil produce 20cu/in at 10 psi.
But its much more efficient to work with 20cu/in at 10 psi, the 200psi at 1cu/in. It takes less time to compress at 10psi with a volume of 20cu/in the compress 200psi at 1cu/in.....
In other words:
If you have 200psi lifting 200lb in one inch taking 1min to reach 200psi, it would take less time and energy to make 10psi with an area of 20in/sq to lift the say weight.
The reason it is more efficient at lower pressure is how hard the compressor needs to work at a higher air pressure. So lower psi the faster the air can be pumped at a great volume.
Just look at a basic house fan, higher volume& lower pressure= low hp....Air compressor at higher pressure& lower volume=geater hp

Tom