Kinematics/Slider Crank Mech. Question(PETE)

[Kevin Shasteen]

Pete...or any other mathmaticians out there....SOS....SOS....SOS.

 

Okay Pete, ever since you dangled the Trig/Algebra in front of my face I have been researching the subject...specifically Algebra.

 

I now understand what the Sin, Cos, Tan, Csc, Sec & Cot represent; yea believe it or not the "Light Bulb" flickered on & I have a firm understanding of their meanings.

 

Now, for my question. What I've been doing is drawing a diagram on paper which represents the piston/rod/crank throw of an engine & determining their relations.

 

The knowns would be the crank throw (1/2 the stroke), length of the connecting rod; for the final known-I'ld pick a * (degree) AFTDC on the crank rotation....say 20* & attempt to find the other variables...length & anlges remaining.

 

My problem is I can get all three lengths but I cant seem to work out the angles of the rod (piston pin center to con.rod center @ crank) both where it connects to the crank journal & also where it connects to the piston pin. In the previous formula you obtained from "Machanics of Machines" you called this angle "Phi".

 

If you do feel like answering this: lets choose a 350 SBC & go with these measurements. If the answer is too involved-just say so & I'll keep hitting the studies till I have another breakdown...oh, I mean a break thru (haha).

 

Rod Length = 5.700

1/2 Crank Stroke = 3.480/2

* ATDC = 20*

 

The geometry (Location of the Triangle) I've been working with has the piston to my left & the crank to my right...on paper that is with the 20* ATDC pointing upward toward the top of the paper.

 

C = Represents Crank Angle

A = Represents Piston/Rod Angle

B = Represents Bottom of Rod to Rod Journal Angle

 

c = leg obviously opposite "C"

a = leg obviously opposite "A"

b = Hypotinuse

 

My problem ocurrs when I approach determining Angle B & C. My formula I have for determining the two is:

 

Tan A = a Sin C/b-a Cos C

Sin B = b Sin A/a

 

Again, if you think the answer is too involved-just say so; no obligation! Just not knowing is driving me crazy!

 

Kevin,

(Yea,Still an Inliner)

 

PS: Gotta go watch the Dallas Stars lose again. Be back in a couple hours.

 

[ May 03, 2001: Message edited by: Kevin Shasteen ]

[Drax240z]

Cool, another Hockey fan in our midst. I won't mention what I think of Dallas. Hehe.

 

Suppose this is your triangle Kevin:

 

 

What we can say about this, if a right angled triangle: (angle a=90*)

 

Pythagoran theorem: A^2 = B^2 + C^2

 

sin © = C/A

cos © = B/A

tan © = C/B

sec © = A/B

csc © = A/C

cot © = B/C

 

And as you've stated (this holds true for any triangle, not just right ones):

 

Law of Sines:

Sin ©/C = Sin (a)/A = Sin (b)/B

 

Law of Cosines:

A^2 = B^2 + C^2 - 2BC Cos (a)

 

And if I am not mistaken, than is all the relationships in a triangle, regarding angles and sides. (obviously not including areas, perimeters, etc)

 

I'm not totally sure I understood what you were trying to find, but it can be done using these formulae if it can be done at all.

[TimZ]

Another couple of tidbits that should help...

 

For any triangle, the sum of the three angles is 180 degrees...

 

a+b+c = 180deg

 

I'm guessing that the problem that you are having has to do with the fact that the triangle that you have described is not a right triangle, so you aren't sure how to apply the trig functions. Another trick that you can use is to split the unruly triangle into two right triangles. Since you know one of the angles, and all of the lengths of the original triangle, you will have one of the angles and the length of the hypotenuse for one of the smaller triangles, which means that you can use the trig functions to get the lengths of its other two sides. Once you have done that, the lengths of the sides of the other triangle are a couple of easy subtractions away, and then you are home free.

[Drax240z]

Oops! That one too. Heh.

[Kevin Shasteen]

Thanks for responding guys, Hey Richard did you happen to catch the Monday Night game between Colorado & LA...that was nothing less than a Braul!

 

What I'm attempting to do is determine the piston's location in the cylinder by knowing the position of the crankshaft. For a quick explanation (quick is a relative term) go into this forum (Miscllaneous Forum) & scroll down till you come to the thread titled "Octane/Comp.Ratio -vs- Cam Selection"; after you read that you will see Pete's response.

 

What I'm attempting to do is map the cranks rotations, in degrees, & come up w/the pistons position w/in the cylinder; this is known as the Kinematic Equation for a Slider Crank Mechanism...so I've recently been told.

 

If you can imagine a cut away of an engine-while its running...you'll see the piston moving up/down in the cylinder as the crank rotates caused by the crank's throw (Stroke) away from the center of the crank.

 

Now picture that crank rotating, piston sliding up/down....and that at any one moment-the engine stops; based on what you know-the length of the rod, the cranks stroke....where would the piston be in the cylinder? That ladies & gentlemen is what I'm attempting to find out.

 

Instead of the random stop of the engine; start the formula while the piston is at TDC & for each crank degree determine the pistons location in the cylinder till you reach BDC. Once you have this you can determine how much Volume is left in the cylinder at any one crankshaft degree. This will aid in the camshaft selection...once you've read the previous mentioned thread "Octane/Comp.Ratio -vs- Cam Selection".

 

So to help me understand the triangle, after drawing the diagram on paper, I then turned it to the left making the piston on its side (horizontally) pointing to the left while the crank is on the right: the centerline from the piston pin to the center of the crank represents the hypotenuse of my triangle. But to determine the location of the piston I have to know the {Angle} of the rod at any given crankshaft position. From Pete's formula this was assigned the name "Phi".

 

Pete's formula was as follows:

 

x = (R+L) - (R cos(theta) + L cos(phi)

 

R = Length of the crank throw (1/2 the stroke)

L = Con Rod Length, center to center

theta = Angle of the crank measured away from TDC (ATDC)

phi = the imaginary centerline from the piston pin center going all the way to the crankshaft journal's center.

 

If I can figure this out/then I wont have to manually mic an engine...BTW-I did manually mic an Olds.Small Block I have & it took me over two days; as I took 4 measurements for every 1*(degree) of crankshaft rotation. Yea it was a pain/but once it's done-its done & I've got it for ever.

 

As you can see by now (I hope) that the triangle created may be a scalene or a right anle; it just depends on where the crankshaft is when you're taking your measurements.

 

My problem is that the angle created by the Connecting Rod (from piston pin to the crank/connecting rod journal) has me stumped.

 

Am I hopeless; or is this something we (Yes I have a mouse in my pocket) can figure out?

 

Kevin,

(Yea,Still an Inliner)

[pparaska]

We can get you going from here. Wow, if you didn't know it, you'd think this was a Homework Forum .

 

Here's a scan from the book:

 

A picture is worth a thousand words, huh?

 

Anyway, the piston is going left-right in that diagram, it's not shown, but it's at point P.

 

Anyway, I think you can see what Tim was talking about, but making right triangles to be able to use trig. That angle is something that you can calculate using the principle that the "leg" along that vertical line below "C" is given by the opposite/adjacent (sin) rule:

That leads to the calculation of phi, as a function of theta, r, and l.

 

I've been working on the spreadsheet and I've recreated the example Kevin put in the other post, but I'm trying to get it to work the inverse problem - given a compression ratio, what intake valve closing (IVC) angle, or theta, gives an X position that's 80%. In other words, leaves 80 percent of the cylinder volume above the piston. I can do it with Excels solver, but not everybody has that loaded. Anyway, I'm about done playing with it and I'll put it up in a few days.

 

HTH,

 

[ May 04, 2001: Message edited by: pparaska ]

 

[ May 04, 2001: Message edited by: pparaska ]

[Kevin Shasteen]

Thanks for humbling me once again Pete (haha); I've noticed a few more symbols in there I wasnt familiar with/looks like my adventure continues-guess I'll have to hit the library a few more times.

 

Thanks for the scan; a piture is definately worth a thousand words....I feel (think) I'm like right on the 'verge' fegureing it all out....it'll just take some thought on my part.

 

Pete; this Spreadsheet-will it be some kind of 'search engine' where one merely fills in the variables & clicks enter for the answer-or are you going to actually input all the answers for one specific (crank/stroke) engine: & will this be on HybridZ or on your personal site?

 

BTW: Barnes & Nobles has "3" books w/the title "Mechanics of Machines" by your Author w/three different copywrite dates. Which copywrite date does your book have?

 

Thanks again Pete for the responses!

 

Kevin,

(Yea,Still an Inliner)

[Drax240z]

Heh, I am taking a course in machine dynamics starting tuesday. I could probably be a lot more help AFTER that course.

 

Lets see, symbols... Forgive me if I am stating the obvious to you, I'm not sure where you are at:

 

I'm sure you got theta and phi Kevin, both being angles.

 

dx/dt is the change in x in terms of t. The derivative of x with respect to t is dx/dt, which is also the velocity, providing x is the position.

 

If you take the derivative of the velocity, or the 2nd derivative of the position, you get acceleration. (d^2x/dt^2)

 

The Ohm sign there is confusing. The capital omega (the one we know as the ohm sign, for resistance) is angular velocity. The lower case omega is the curvy w. For some reason they aren't consistant with their usage, I'm not sure why.

 

Alpha is the angular acceleration.

 

This should be all covered in my course in machine dynamics, as I said. Unfortunately I don't know a lot about it yet.

[clint78z]

I find the easiest way to explain derivatives in calculus is to go back to a simple graph that most people know. Think of a car in a 1/4 mile. x is the distance from start and t is the timeon the bottom of the graph.

 

x=dist | ft

t=time | sec

 

When doing a deravative on the graph what you are really doing is checking the velocity of the car at each time.

 

So the second graph after you take the deravative is

 

v= velocity | ft/sec

t= time | sec

 

Third graph

 

a=acceleration | ft/sec^2

t=time | sec

 

Velocity, acceleration are all related back to the distance and time. Calculus just uses fancy math to find velocity and acceleration.

 

Hopefulley no one was to confused, mabey someone got a light bulb after that.

[Kevin Shasteen]

So, t = time while v = velocity; I was wondering where those two came into play. Also, what are the three dots forming a triangle & what do they represent?

 

If all I want to know is position; do I have to worry about velocity?

 

How bout we use the 350 SBC & work this problem thru? One step at a time? One I'm shown how to work a math problem I usually dont have any problem picking it up; its obvious if I too go on this alone it'll take me forever...as it's taken my a couple weeks just to understand Sin, Cos, ect , ect!

 

BTW-I found software that does just what I'm looking for...determines velocity, position & you can even change the length of the rod & the stroke; all for only $15. The only drawback is that its in MacIntosh format Doh! I'm doing good just to learn the IBM/Microsoft stuff...much less invest in a Mac set up!

 

Thanks for the pointers everyone.

 

Kevin,

(Yea,Still an Inliner)

[johnc]

Uuuhhhh... there are a bunch of software programs out there that figure this stuff out. Here's a link to a lot of stuff: http://roadsters.com/software/#software

[pparaska]

John, thanks for that link. Great index of software there. However, the ones that seem to be able to do any of what were talking about are commercial, and the free stuff I looked through briefly didn't include the calculations I'm looking for.

 

Kevin, I'm not sure what form the spreadsheet will be in. But I'll try to make it user-friendly. I'll get you the particulars on that book this weekend.

[Drax240z]

The three little dots are used often for 'therefore' in mathematics, engineering, physics, etc. Basically the step before you've shown something, and therefore you can derive something else from it.