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Or Seth, either one
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Discussion Starter #1 (Edited)
Trying to understand a few things. First is a Radian.

In reading the Wikipedia page on it...

The radius of a circle, in engine terms would be half the stroke, as the connecting rod journal axis creates a revolution, it makes a circle. A 4" stroke will create a 4" circle and the radius of that circle is 2". A Radian is a portion of the circles circumference equal to the radius. So, if the rod journal were at top dead center and you rotated the crank one Radian, the center axis of the journal would move along circular curve of the circumference 2". In a sense, creating a triangle 2" from the center of the crank to the rod journal at TDC, 2" along the circumference (of the circle the rod journal travels), and 2" from that point back to the center of the crankshaft.

I'm trying to figure out how many Radian the connecting rod journals axis travels in one stroke, from TDC to BDC. The best I make of it, and it's becoming more clear as I write this, is there are 3.14 (Pi) Radian in a semi-circle and 6.28 (two Pi) Radian in a circle. I should be able to check this against the equation for finding the circumference of a circle.

The circumference of a circle = Pi x Diameter = 3.14 x 4 = 12.56" circumference. Half of that circle (180°) = 6.28".

One Radian of a 4" circle = 2"... 3.14 Radians x 2" = 6.28". AWESOME, that checks out for a half circle, or one stroke from TDC to BDC.

I think I just wrapped my mind around that part.

What I'm ultimately trying to understand is the equation for finding peak piston velocity, which calls for replacing a constant known angle in a triangle into a known angular velocity, or a moving angle. Angular velocity - Wikipedia, the free encyclopedia The three points of this triangle being the wrist pin axis, rotating axis of the crankshaft, and the axis of the connecting rod journal. The known sides of that triangle are from the crank axis to the rod journal axis (half of stroke) and the connecting rod length. The known angle is degrees ATDC.

My best understanding at this point, is a known constant angle in the "Law of Cosines" formula (used to find unknown angles/legnths with known angles/legnths) the equation is replaced with a known angular velocity to find the unknown angular velocity. So if we know that the rod journal sweeps from 0° to 180° in a certain time, or a certain number of repetitions in a given time (RPM), we can calculate the speed of the wrist pin axis. That's what I understand on a theory basis, but haven't 100% wrapped my mind around it to the point of full understanding.

Any input you have to help me further understand would be deeply appreciated.
 

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Trying to understand a few things. First is a Radian.

In reading the Wikipedia page on it...

The radius of a circle, in engine terms would be half the stroke, as the connecting rod journal axis creates a revolution, it makes a circle. A 4" stroke will create a 4" circle and the radius of that circle is 2". A Radian is a portion of the circles circumference equal to the radius. So, if the rod journal were at top dead center and you rotated the crank one Radian, the center axis of the journal would move along circular curve of the circumference 2". In a sense, creating a triangle 2" from the center of the crank to the rod journal at TDC, 2" along the circumference (of the circle the rod journal travels), and 2" from that point back to the center of the crankshaft.

I'm trying to figure out how many Radian the connecting rod journals axis travels in one stroke, from TDC to BDC. The best I make of it, and it's becoming more clear as I write this, is there are 3.14 (Pi) Radian in a semi-circle and 6.28 (two Pi) Radian in a circle. I should be able to check this against the equation for finding the circumference of a circle.

The circumference of a circle = Pi x Diameter = 3.14 x 4 = 12.56" circumference. Half of that circle (180°) = 6.28".

One Radian of a 4" circle = 2"... 3.14 Radians x 2" = 6.28". AWESOME, that checks out for a half circle, or one stroke from TDC to BDC.

I think I just wrapped my mind around that part.

What I'm ultimately trying to understand is the equation for finding peak piston velocity, which calls for replacing a constant known angle in a triangle into a known angular velocity, or a moving angle. Angular velocity - Wikipedia, the free encyclopedia The three points of this triangle being the wrist pin axis, rotating axis of the crankshaft, and the axis of the connecting rod journal. The known sides of that triangle are from the crank axis to the rod journal axis (half of stroke) and the connecting rod length. The known angle is degrees ATDC.

My best understanding at this point, is a known constant angle in the "Law of Cosines" formula (used to find unknown angles/legnths with known angles/legnths) the equation is replaced with a known angular velocity to find the unknown angular velocity. So if we know that the rod journal sweeps from 0° to 180° in a certain time, or a certain number of repetitions in a given time (RPM), we can calculate the speed of the wrist pin axis. That's what I understand on a theory basis, but haven't 100% wrapped my mind around it to the point of full understanding.

Any input you have to help me further understand would be deeply appreciated.
I do it the easy way. I look up the piston travel per degree on a chart and look for the max positon movement.



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Or Seth, either one
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Discussion Starter #3 (Edited)
I do it the easy way. I look up the piston travel per degree on a chart and look for the max positon movement.
I appreciate the easy way. Nothing wrong with it at all. There's even online calculators.

Just outta curiosity, are these chart's you speak of on line or hard copy? Are there different charts for different rod ratios? Might be helpful in understanding everything to have a visual. Gotta link if they're on-line?

Also, when you reference the chart, I assume it gives you the crank angle at which it occurs, but how do you translate that into a measurable speed in fps?
 

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I appreciate the easy way. Nothing wrong with it at all. There's even online calculators.

Just outta curiosity, are these chart's you speak of on line or hard copy? Are there different charts for different rod ratios? Might be helpful in understanding everything to have a visual. Gotta link if they're on-line?

Also, when you reference the chart, I assume it gives you the crank angle at which it occurs, but how do you translate that into a measurable speed in fps?
Yes they are out there in different rod ratios. In have some in hard copy and some are online. They are all available in heard copy if I hit "PRINT":)sphss

Of course they give you the crank angle, they are done in piston movement PER DEGREE!!!!!!

I guess if you need to know the exact psiton speed, which ever changing thru the entire stroke, you could figure it out by determining how long it takes a crankshaft to rotate 1 degree and how far the psiton moved in that amount of time.
But he piston is pretty much in a state of acceleration and de-acceleration.

Like trying to determine how fast a fueler is going at any given foot, or inch, or even 1/4 inch, or ..... of the track or the exact location of an electron in its obit.



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Not a chart, but it illustrates that not much change in speed/acceleration takes place over much of the full rotation. The majority of the change is at TDC and BDC and they basically swap accleration loss or gain. You are slowing it down on one end of the stroke and speeding it up on the other.

As the piston losses velocity up to and away from TDC with a longer rod, it gains velocity at BDC. Opposite when the rod gets shorter.








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steelcomp was here
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Trying to understand a few things. First is a Radian.

In reading the Wikipedia page on it...

The radius of a circle, in engine terms would be half the stroke, as the connecting rod journal axis creates a revolution, it makes a circle. A 4" stroke will create a 4" circle and the radius of that circle is 2". A Radian is a portion of the circles circumference equal to the radius. So, if the rod journal were at top dead center and you rotated the crank one Radian, the center axis of the journal would move along circular curve of the circumference 2". In a sense, creating a triangle 2" from the center of the crank to the rod journal at TDC, 2" along the circumference (of the circle the rod journal travels), and 2" from that point back to the center of the crankshaft.

I'm trying to figure out how many Radian the connecting rod journals axis travels in one stroke, from TDC to BDC. The best I make of it, and it's becoming more clear as I write this, is there are 3.14 (Pi) Radian in a semi-circle and 6.28 (two Pi) Radian in a circle. I should be able to check this against the equation for finding the circumference of a circle.

The circumference of a circle = Pi x Diameter = 3.14 x 4 = 12.56" circumference. Half of that circle (180°) = 6.28".

One Radian of a 4" circle = 2"... 3.14 Radians x 2" = 6.28". AWESOME, that checks out for a half circle, or one stroke from TDC to BDC.

I think I just wrapped my mind around that part.

What I'm ultimately trying to understand is the equation for finding peak piston velocity, which calls for replacing a constant known angle in a triangle into a known angular velocity, or a moving angle. Angular velocity - Wikipedia, the free encyclopedia The three points of this triangle being the wrist pin axis, rotating axis of the crankshaft, and the axis of the connecting rod journal. The known sides of that triangle are from the crank axis to the rod journal axis (half of stroke) and the connecting rod length. The known angle is degrees ATDC.

My best understanding at this point, is a known constant angle in the "Law of Cosines" formula (used to find unknown angles/legnths with known angles/legnths) the equation is replaced with a known angular velocity to find the unknown angular velocity. So if we know that the rod journal sweeps from 0° to 180° in a certain time, or a certain number of repetitions in a given time (RPM), we can calculate the speed of the wrist pin axis. That's what I understand on a theory basis, but haven't 100% wrapped my mind around it to the point of full understanding.

Any input you have to help me further understand would be deeply appreciated.
Ha! Got ya thinkin, eh. ;)
 

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Or Seth, either one
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Discussion Starter #8
Not a chart, but it illustrates that not much change in speed/acceleration takes place over much of the full rotation. The majority of the change is at TDC and BDC and they basically swap accleration loss or gain. You are slowing it down on one end of the stroke and speeding it up on the other.

As the piston losses velocity up to and away from TDC with a longer rod, it gains velocity at BDC. Opposite when the rod gets shorter.
Very constructive post Bob. Thanks!

I've seen this particular one in my research. What the graph doesn't mention as posted here, is the stroke and rpm. However, what it does show is (looking at the 2:1 rod ratio) peak piston speed reaches about 7500 feet per minute, or 125 feet per second, or 1500 inches per second.

If the piston is in a 4.5" bore, it's surface area is... (radius x radius x Pi) 4.5 diameter/2= a radius of 2.25 x 2.25 radius = 5.0625 x 3.14 = 15.89625 sq. inches. (Round to 15.90)

If 15.90 square inches is moving at a rate of 1500 inches per second, it is drawing at a rate 23850 cubic inches per second. Divide by the number of cubic inches in a cubic foot,1728 and you get 13.80 cubic feet per second. Multiply by 60 gives you maximum CFM draw of 828cfm.

Again, we don't know what rpm or stroke is in the graph...

Does that mean anything that's worth anything? Dunno.

I still want to learn and completely understand the equation used to find peak piston velocity.

Anyone? A little help here?
 

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Or Seth, either one
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Discussion Starter #9 (Edited)
BTW Budweiser, its not physics. Its geometry, trig and alittle calculus thrown in. But its not physics.
Calculating the angles would be geometry, however once those angles are in motion, I'm pretty sure it's physics.

But hey! I never made it through geometry, so WTF do I know?

What ever it is... I want to understand it. Call it knitting for all I care. :happy:

The quote below is from: Wikipedia. What is a Radian?

Use in physics

The radian is widely used in physics when angular measurements are required. For example, angular velocity is typically measured in radians per second (rad/s). One revolution per second is equal to 2π radians per second.
Similarly, angular acceleration is often measured in radians per second per second (rad/s2).
For the purpose of dimensional analysis, the units are s−1 and s−2 respectively.
Likewise, the phase difference of two waves can also be measured in radians. For example, if the phase difference of two waves is (k·2π) radians, where k is an integer, they are considered in phase, whilst if the phase difference of two waves is (k·2π + π), where k is an integer, they are considered in antiphase.
See why I'm having trouble??? :)sphss
 

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After looking at graph charts for a BBC 4" stroke with rods at 6.135, 6.385, 6.535 on the same graph, and not being able to see the 3 lines as anything but one thick line, I really don't get to overly exciting about the location, or even the velocity of the piston at any given point. I
I like long rods for 4 reasons. It MAY take a little load strain off the piston and pin at the top of the stroke, better ring seal due to better rod angle. more time for the cylinder pressure to rise before decending down the bore, possibly allowing for less timing, but also possibly adding to the chance of detontation, and less rod angularity there for less friction on the piston skirt.

I could care less about any of the other aspects of long rods. I know Steel has thoughts about the other things regarding longs rods as do others. It one of those things when doing any dyno testing, what exactly was the reason for any gains or losses.

Guys like sonny Leonard could care less about rod length. He simply wants rod long enough to get the piston pin to clear the counter weights. Rod length = stroke +2" is pretty normal for him and amny others that build huge CID engines.
One of Scott Shafiroffs best sellers is a 598 with a 4.5 crank and 6.535 rod in a low deck. When you look at the pin CH, short ass skirt and the rod angle, it is a major loser for rings seal and friction in my mind, but he sells the shit out of them.



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Or Seth, either one
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Discussion Starter #13 (Edited)
The graph you (GN7) posted is from one of the pages I've been referencing for the calculation.

ftlracing.com

Walk me through it, please.

Anyone?
 

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Not sure what exactly needs explaining. Its like I said in an earlier post. The shorter rod is quicker up into the chamber and out, both acclerate at the same rate at some point in the revolution, then the short rod is a little lower thru the bottom of the stroke.

There is something you need to pay atttention to when looking at graphs like this. The one you linked was between a 5" and 7" rod on a 3.5 stroke. Ever seen a 7" rod on a 3.50 stroke? I have built a 6 rod on a 3.00 stroke before. But the point I am making is that the comparsion is considerably larger than you will normally see when looking at different possible combos.

lLook at this graph. A 4" rod compared to a 12" rod. that would be like comparing a stock 6.135 BBC rod to a 18 3/8" rod. Not going to happen. the graphs like this are great because it allows you to see that something is happening to the piston velocity. But in the real world, its not that big of a difference. Like I said, I have a graph for 3 rod lengths used on 4" BBCs and the lines are right on top of each other and look like in thick line.





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Discussion Starter #15
Ha! Got ya thinkin, eh. ;)
Didn't see this post earlier.

Uh, yeah. Mind spinning. Can't sleep. Digesting all this like a good steak and potato's meal.

"Thinkin" is an understatement! :))THumbsUp
 

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Not a chart, but it illustrates that not much change in speed/acceleration takes place over much of the full rotation. The majority of the change is at TDC and BDC and they basically swap accleration loss or gain. You are slowing it down on one end of the stroke and speeding it up on the other.

As the piston losses velocity up to and away from TDC with a longer rod, it gains velocity at BDC. Opposite when the rod gets shorter.
I think the chart isn't labled really well. I had to look at it a bit to make sense out of it. It's labled in velocity and is correct based on crank angle, but actually shows peak piston acceleration at the top and bottom of each stroke and is zero at mid stroke. If you take the first integral of acceleration, you get velocity but there is a 90* phase shift associated with the calculation. If you work out the math, peak piston velocity is a the zero crossing of the sinusoidial representation of acceleration you posted. Acceleration is at it's peak on either end of the sine wave and if it's plotted properly, the end points are the length of the stroke. It's much easier to explain if I had a tablet and could draw it out as I explained it. For the ME's in the bunch, it's basic Jeffcott rotor theory for the relationship between displacement (stroke), piston speed (velocity in inches per second) and acceleration (velocity squared). All are either integrals or derivitaves of each other (depending on which way your changeing parameters).
The use of radians makes the math cleaner for calculations.
 

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Or Seth, either one
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Discussion Starter #17 (Edited)
I think the chart isn't labled really well. I had to look at it a bit to make sense out of it. It's labled in velocity and is correct based on crank angle, but actually shows peak piston acceleration at the top and bottom of each stroke and is zero at mid stroke. If you take the first integral of acceleration, you get velocity but there is a 90* phase shift associated with the calculation. If you work out the math, peak piston velocity is a the zero crossing of the sinusoidial representation of acceleration you posted. Acceleration is at it's peak on either end of the sine wave and if it's plotted properly, the end points are the length of the stroke. It's much easier to explain if I had a tablet and could draw it out as I explained it. For the ME's in the bunch, it's basic Jeffcott rotor theory for the relationship between displacement (stroke), piston speed (velocity in inches per second) and acceleration (velocity squared). All are either integrals or derivitaves of each other (depending on which way your changeing parameters).
The use of radians makes the math cleaner for calculations.
I was hoping you would pop your head in. Not sure there are too many ME's in here.

I think you may have misread the graph. It represents the velocity (or speed) of the piston through one complete crankshaft revolution. On the left it TDC, the center BDC, and the right is TDC. It's a graph of velocity, not acceleration. Although, the slope of the graph is an indication of acceleration.

So, at 45° ATDC the piston is traveling 5000fpm. Somewhere around 90° it maxes at about 7500fpm... And at 180° (middle of graph) the piston is at BDC... 0fpm. Then it starts it's way up...

More questions in a few. Gotta compile.
 

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I was hoping you would pop your head in. Not sure there are too many ME's in here.

I think you may have misread the graph. It represents the velocity (or speed) of the piston through one complete crankshaft revolution. On the left it TDC, the center BDC, and the right is TDC. It's a graph of velocity, not acceleration. Although, the slope of the graph is an indication of acceleration.

So, at 45° ATDC the piston is traveling 5000fpm. Somewhere around 90° it maxes at about 7500fpm... And at 180° (middle of graph) the piston is at BDC... 0fpm. Then it starts it's way up...

More questions in a few. Gotta compile.
You see it correctly. The piston is at zero velocity at both ends of the graph(TDC) and mid graph(BDC) and max velocity at roughly 90* both up and down.
did you happen ito notice that it took a comparison between a 5 inch rod and a 7 inch rod to make that much difference in the graph. when you look at a piston dispalcement graph, the lines get even closer with the same ratios.



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Not a chart, but it illustrates that not much change in speed/acceleration takes place over much of the full rotation. The majority of the change is at TDC and BDC and they basically swap accleration loss or gain. You are slowing it down on one end of the stroke and speeding it up on the other.

As the piston losses velocity up to and away from TDC with a longer rod, it gains velocity at BDC. Opposite when the rod gets shorter.



Either I am slightly confused or the charts aren't exactly agreeing with you.

A LONGER rod will have a slower approach and departure from BOTH TDC and BDC.

Whereas a shorter rod will have a quicker approach and departure from both TDC and BDC.

The colored lines on both charts agree, sort of a simple syne wave.

If either rod had a differing speed at either TDC or BDC it would be clearly shown that the wave form would be slightly biased up/down from their current position and actually crossing the waveform of the 'other' rod.

Correct?

Is this a test?

Did I pass or fail?

I need to look up some more stuff for study.
 
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