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/C...nk%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?

 

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!

(P.S. Am I correct? )

 

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/C...nk%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?