Geometry Lesson - Dwell Time

A longer Rod influences the piston to dwell a bit longer at TDC than a shorter rod would and conversely, to dwell somewhat less at BDC. This is another area where people often get the information backwards.

IE: piston 40 degrees BBDC to 40 degrees ABDC , compared with 40 degrees BDTC to 40 degrees ATDC

How / why does the dwell change between TDC and BDC? We are assuming
a steady crank rotation correct?

why would it not have the same dwell time at BDC? doesnt make sense to me... :confused:

Me either, unless it somehow has to do with stretching of the rod at TDC or something?

Why do you even care what it's doing at BDC? Dwell time at the top affects how the fuel is burnt.

Because a longer dwell at BDC will allow a later closing intake valve (IVC) to increase effective
compression.

More charge into the compression and power stroke for more torque.

Because a longer dwell at BDC will allow a later closing intake valve (IVC) to increase effective
compression.

More charge into the compression and power stroke for more torque.

I do know this much, unless you just want to know to know which is why Im curious. Youre talking an amount of time that is not even remotely worth worrying about in the case of 5.7-6" rods.

To know...is all I want to know.

I agree based on what I've read. But allowing the intake valve to close later
seems to be very important.

The next question after figuring out my first is why a shorter rod has more
duration at BDC than a longer rod?

The plot thickens!

OK, I think I have the short vs. long rod dwell figured out.

The acceleration time above 20 degrees before and after TDC/BDC is slower with a shorter rod.

That's easy to understand.

That allows us to close the intake valve a little later (~ 2 - 4 degrees according
to ISKY cams).

As far as the first question, I'm still drawing a blank.

OK, I think I have the short vs. long rod dwell figured out.

The acceleration time above 20 degrees before and after TDC/BDC is slower with a shorter rod.

That's easy to understand.

That allows us to close the intake valve a little later (~ 2 - 4 degrees according
to ISKY cams).

As far as the first question, I'm still drawing a blank.

Homework assingment time Tino:

Draw a schematic diagram of crank and rod. Crank pin turns thru 360 degrees, and large end of rod follows the circle. Small end of the rod (and therefore piston) moves linearally along an axis which passes thru the center of rotation of the crank (center of main bearing). The angle between the rod (center-to-center of the big end and small end) and the rod journal changes with rotation , of course. The trig is fairlyt straight forward, and doing a spreadsheet which calculates the distance of the small end of the rod from the crank centerline for every degree of rotation should be realatively easy to write. Make the stroke (or half the stroke, the "throw") and the rod length a variable, and you can "plug 'n crank" for any combination. Look at the last 10 degrees or so either side of TDC and BDC and you'll get your answers.

I'd rather teach you to fish than give you the fish.

It you have problem with trig and algebra, you 'll have to find someone who's already done this. "SEARCH" is your friend.

Way ahead of you Jon.

I appreciate the hints rather than the answer. I know I've been lazy in the
past, but lately, Google and I have spent many hours together.

There is an article which is touching on what you have just posted. Still reading
through it.

Thanks for the post!

http://yarchive.net/car/index.html

http://yarchive.net/metal/piston_acceleration.html

http://groups-beta.google.com/group/rec.crafts.metalworking/browse_thread/thread/95ca6f1759484c7a/65b473fe43e32c17?q=group:rec.crafts.metalworking+i nsubject:acceleration+insubject:maximum+insubject: piston&rnum=1#65b473fe43e32c17

Here's the first part of the homework assignment.

It's not quite to scale, but it's a visual for those wanting to follow along:

http://gmthunder.com/tino/rodangle.jpg

Ok, I feel really stupid right now...trying to write up this excel worksheet to plug in stroke, rod, and compression heights and based on degrees of crank rotation output the piston height above crankpin centerline, and rod distance left to right of crankpin centerline.

Im just working on the piston height right now and for some reason my trig isnt coming back to me. If you know the angle at the crankpin (obvoiusly) say 75*, and you know the adjacent side (say 2"), and you know the opposite side (say 6"). What trig function would you use to figure out the last side?

And would you have to change the funtion for a different group of degrees, say after 180 or anything?

I'm in the same place you are.

The search turned up very little with respect to our needs.

I'm thinking of using Pythagorean's Theorum to find the piston pin centerline
above the crankshaft centerline.

From there we can assume a 90 degree angle along the cylinder bore vertical.

EDit: Scrath that!

I think I've got it!!!!

Without the math, here's the visual:

With the rod and crank arm at 90 ATDC, the rod is creating the biggest angle
at the piston wrist pin.

After the large rod end goes past the centerline, the piston is still moving closer
to the crank centerline.

The distance of the piston pin from centerline at 0' TDC is 1/2 stroke + rod length.

6.0" + (3.48 /2) = 7.74"

The distance of the piston pin from centerline at BDC is 1/2 stroke - rod length (absolute).

6.0" - (3.48 /2) = 4.26"

Thinking out loud....

angle of rod at TDC vs BDC +/- 10 degrees...rod and crank assembly is virtually 3.48" shorter at BDC.

That would explain the difference in time to move the piston when it's way
up in the bore, as opposed to far down in the bore.

Are we getting it Jon?

Math to follow tomorrow night.

revised diagram to show crank arm (green rectangle)

http://gmthunder.com/tino/rodangle2.jpg

Getting closer......

I'm sure he'll say something in the AM.

Bret

Im just working on the piston height right now and for some reason my trig isnt coming back to me. If you know the angle at the crankpin (obvoiusly) say 75*, and you know the adjacent side (say 2"), and you know the opposite side (say 6"). What trig function would you use to figure out the last side?

And would you have to change the funtion for a different group of degrees, say after 180 or anything?

Signs may change as the crank moves from quadrant (0 to 90 deg or "12 to 3 o'clock") to quadrant (90-180*, or "3 to 6 o'clock"). etc.

SIN = Opposite/hypotenuse (SOPH)
COS = adjacent / hypotenuse (CAP)
TAN = opposite/adjacent (TOAD)

If you remember SOPH, CAP & TOAD you can always derive what you need. That takes less brainram than memorizing the entire formula.

Well I think I got it in Excel...

https://webspace.utexas.edu/srl264/Crank/Crank%20angle.xls

I think I may have messed up somewhere, but I think I got it. I separated it by 1*, so its kinda long. Let me know how I did and if I messed up, where my calculations were off.

-Stu

anyone? is my spreadsheet accurate?

anyone? is my spreadsheet accurate?

How about labelling what things are. It's confusing to me.

Alright I modified it and actually think I fixed my original problem (i was having abs value issues due to sin being an odd function).

I fixed it now and changed the piston height to express a comparison between 6, 5.7, 4, and 8" rods and how they would look on a 9" deck SBC with 1, 1.3, 3, and -1" pistons. The 4 & 8" rods are not possible on a 4" stroke, but are simply for comparison purposes because the idea is the same.

The crank degrees are in the 1st column and express degrees rotation starting at TDC. Every thing else should be self explanitory.

-Stu

If you wanted to modify this to express your motor, you can modify this

(=SQRT(6^2-(SIN(A2*PI()/180))^2*2^2)+COS(A2*PI()/180)*2+1 )

formula and change the bold 2s and put in half your stroke and the bolt italic 1 can be changed accordingly to match your deck height. Then copy and paste that down the column to find your piston height at a given crank rotation.

I think you have some errors in your math yet?

If you have half the stroke (1.54"), plus the rod (6.0"), the max center to center would be 7.74" at TDC.

Here are 2 points I've calculated using Pythagorean's Theorum, and two others
with simple addition and subration:

Crank centerline to Small Rod end center:

0 TDC = 7.7400"
90 ATDC = 5.7422"
0 BDC = 4.2600"
90 ABDC = 5.7422"

c^2= a^2 + b^2

6.0^2=1.74^2 + b^2

36 = 3.0276 + b^2

36 - 3.0276 = 3.0276 - 3.0276 + b^2

32.9724 = b^2

Sq. root of 32.9724

5.7422 = b

b = vertical of cylinder bore.

Now I'll go back and figure each degree of rotation using those 4 points as
checksums.

(Jon, you should be getting paid for this!)

Pictures are worth a thousand words, and I totally understand what is happening
now that I'm focusing on the crank throw AND rod angle to determine dwell.

I used diagrams and long math to figure this out instead of on-line calculators;
not to be difficult, but to understand what is happening.

After trying the long math version of this, I'll trust that the other angles fall
into place.

Here's a couple of pics:

http://gmthunder.com/tino/homework.jpg
http://gmthunder.com/tino/rodtrig.jpg

I drew an imaginary line to break the obtuse triangle into two right angle triangles.

From there, I had the crank throw and rod length, plus one 90 degree angle
and a given crank rotation angle.

Using the TAN, COS and Pythagorean's Theorum, I cam up with 7.73148" at 5 degrees ATDC.

Thank you Jon (and others) for helping me through this. Just to let you know,
there are people that take these threads to heart and dive into the math! :bow:

(P.S. Am I correct? :D )

My math was dead on, I just used a 4" crank and added in piston height to make a 9" deck height for easy comparison between different rod lengths. I just copied and pasted my formula into the eqn... look how mine came out to 7.73" (I actuallu calculated 7.73146 on my ti-89... looks like your calc is 2/10000 off, lol)

BTW, pythagorean is not the easiest way to go about solving for that. You would be better off with a combination of sine and cosine laws. It solves it rather easily. I also threw compiled a formula that calculates the wrist pin height on a 6" rod and 3.48 stroke from TDC all the way around in intervals of 1*.

I also threw in there piston speeds at various RPMs at a given crank position. Im not 100% sure about the units on the velocity, but im pretty sure its right.

Yea, ive been bored and had free time.

https://webspace.utexas.edu/srl264/Crank/Crank%20angle.xls

^--- could prove quite useful to many people with stock bottom ends.

I realize I could have used the TAN, COS, SIN using two sides and an angle,
but I wanted to stay away from calculator functions as much as possible. I'm just a little stubborn that way. I love pain!

As for your spreadsheet, I'm still having trouble understanding your numbers.

All the values seem nearly equal across the board even when there is a difference
of 4" in rod length (columns D and E)?

Maybe I'm not understanding the legend?