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Old 03-16-2017, 03:29 PM   #21 (permalink)
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Has OP (or anyone else, for that matter) found a way to overcome the tendency of all of the sprung weight of this piston/spring conrod contraption to resist the continual changes in velocity due to moving up and down within its cylinder?

That continually varying momentum of the sprung mass, interacting with the elastic nature of the spring itself, is going to set up for some rather undesirable engine speed-dependent harmonic motion.

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Old 03-16-2017, 05:29 PM   #22 (permalink)
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Originally Posted by ksa8907 View Post
Ideally, assuming this idea can work, i think anticipating the spring compression and then increasing the compression ratio back up to what is common on todays engines could mean big power numbers. You would effectively be getting a longer stroke out of any given engine.

Another effect, as the piston is decelerating toward bdc on the intake stroke, it may also compress the spring at higher engine speeds allowing for a larger effective displacement than static.
...maybe not since the spring would definitely be compressed on tdc.
The compression ratio shouldn't change. The spring should be stiff enough to only start compressing after fuel burning has well progressed. I don't see the need to increase power in today's engines, only reduce wasted heat(increase efficiency) to get more energy to the wheels and less to the exhaust and radiator.

From what I have learned about engines you need a certain compression to get the air/ air fuel mixture hot enough. After that it will ignite(by itself or sparkplug) and burn by itself. After igniting it releases heat which raises temps and pressure more. At a certain point the combustion will be hot enough so there is no combustion gain raising the temps more.

The high temps get absorbed by the cylinder walls, head and piston. Which is waste heat to radiator eventually. If the peak pressure and thus temps is lower, the difference between engine temp and combustion temps mean less heat will get absorbed by engine.

Also T Vago:
The spring should be stiff enough not to compress at BDC, only closer to combustion peaks. If it does compress at BDC I think the engine will have problems during combustion to burn the fuel. Someone who knows how to work these things out will have to enlighten us.
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Old 03-16-2017, 06:01 PM   #23 (permalink)
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I've always had a issues with spring problems in physics classes and those friction-less ladder problems too.

Guess you can tag me out on this topic.
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Old 03-16-2017, 07:59 PM   #24 (permalink)
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So at a 30:1 compression ratio (assuming temperature of the compressed gas stays at 60f which ot won't), the spring will need to be able to resist (425 psi x (piston area)in2) roughly 4500lbs depending on piston size and the effective dynamic compression ratio.

Is that math right? Seems high.

If that is accurate, you're certainly not using a mechanical spring.


I re-ran some numbers using 17.3:1 compression of a cummins 6.7 and 15 psig of boost, roughly 2700 lbs of force.
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Old 03-16-2017, 11:41 PM   #25 (permalink)
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Even though your intention doesn't sound bad, it's neither simple or practical at all.
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Old 03-17-2017, 03:07 AM   #26 (permalink)
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Quote:
Originally Posted by ksa8907 View Post
So at a 30:1 compression ratio (assuming temperature of the compressed gas stays at 60f which ot won't), the spring will need to be able to resist (425 psi x (piston area)in2) roughly 4500lbs depending on piston size and the effective dynamic compression ratio.

Is that math right? Seems high.

If that is accurate, you're certainly not using a mechanical spring.


I re-ran some numbers using 17.3:1 compression of a cummins 6.7 and 15 psig of boost, roughly 2700 lbs of force.
Thanks, this is what I was looking for. Some math that says it's possible or not. I can't say if it is correct or not as I don't know how to calculate it.

What bothers me is that a small starter motor is able to turn the engine against the compression and start it. Maybe it's the gearing that makes it possible.

Maybe there is someone that can point us to the correct formulas, etc. to calculate it.
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Old 03-17-2017, 09:52 AM   #27 (permalink)
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Quote:
Originally Posted by DieselJan View Post
What bothers me is that a small starter motor is able to turn the engine against the compression and start it. Maybe it's the gearing that makes it possible.

Maybe there is someone that can point us to the correct formulas, etc. to calculate it.
Thanks
If the resistance due to compression was anywhere near the force of expansion, the engine wouldn't turn at all.
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Old 03-17-2017, 10:56 AM   #28 (permalink)
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Quote:
Originally Posted by ksa8907 View Post
........I re-ran some numbers using 17.3:1 compression of a cummins 6.7 and 15 psig of boost, roughly 2700 lbs of force.
That's a really high compression ratio and those are large cylinders.

Still a ton = 2,000 lbs and that's what I posted earlier, over a ton of force.

Your math matches.

Next great idea.................?
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Old 03-17-2017, 11:08 AM   #29 (permalink)
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Seems like excess complexity to me, especially for a critical component that has many harsh demands placed upon it.

Consider how such a part would wear too.

I've read about spring-loaded detonation dampers located in the cylinder head which would be orders of magnitude simpler.
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Old 03-17-2017, 12:34 PM   #30 (permalink)
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Magnets:they really do not like heat...

That amount of force, your springs are roughly the size of the springs in your suspension.

Also, when you fondle a spring, it goes wobbly wobbly, that could be deteimental in an engine that has variable rpms. You should look at tuning mass spring damper systems (if you have friends that are good in electronics, it is the same as RLC {capacitance, inductance and resistance} circuits).

Also at those speeds and forces I doubt that your spring will be linear (which may or may not be to your advantage).

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