Quick \"Trig\" question.... OK, I was bored in a meeting today, and I was sketching out a simple diagram for a linked suspension.... /forums/images/graemlins/waytogo.gif Assume you had a front suspension link that was 36" long and was completely parallel to the ground. The link travels up a maximum of 7", and down a maximum of 7" (This is a hypothetical 14" shock set up for equal travel in both directions). How do I calculate the distance that the wheel has moved "back" at full stuff and full droop? It's travelling in an arc, so there is a slight backward movement as the suspension cycles....I drew it on graph paper, and looked for some "triangles" that I could superimpose on the sketch that would correctly represent the value I'm trying to solve for..... but couldn't come up with one. I know that the number I'm looking for is probably only about 1" (or slightly less) from my scaled drawing, but I'd like to know how to really calculate it. It's the kind of thing that might eventually be useful to add to ExcelCAD 2.1??? /forums/images/graemlins/thinking.gif

Re: Quick \"Trig\" question.... Simple triangles man, no need for all the fancy stuff. Just calculate the base length of a right triangle having a hypotenuse equal to the length of an arm and the vertical height equal to the offset of the link end (7” in your query). So: a^2 + b^2 = c^2 a^2 + 7^2 = 36^2 a^2 = 36^2 – 7^2 a = SQRT( 36^2 – 7^2 ) So the offset will be 36 – a 36 – 35.313 The answer is 0.687

Re: Quick \"Trig\" question.... [ QUOTE ] Simple triangles man, no need for all the fancy stuff. Just calculate the base length of a right triangle having a hypotenuse equal to the length of an arm and the vertical height equal to the offset of the link end (7” in your query). So: a^2 + b^2 = c^2 a^2 + 7^2 = 36^2 a^2 = 36^2 – 7^2 a = SQRT( 36^2 – 7^2 ) So the offset will be 36 – a 36 – 35.313 The answer is 0.687 [/ QUOTE ] /forums/images/graemlins/bow.gif ...so simple, yet not. .

Re: Quick \"Trig\" question.... Triangles and the Pythagorean Theorem will get you to the solution of most problems like this. There are often simpler plug in formulas but those are harder to remember and apply correctly. For example, even if you didn't start from a horizontal link, you could just calculate this twice to get the base length for the starting point and the target point, then just calculate get the difference to figure out how far off it will move.

Re: Quick \"Trig\" question.... You know.....as soon as you told me, I was like "Ohhhhhh yeah, that's it." I don't even want to admit all the crazy ways I was trying to apply SIN, COS, etc... the fact that the link traveled in an arc had me obsessed with finding some sort of complicated triginometric solution. /forums/images/graemlins/grin.gif My 6th grade math teacher always told me that Pythagoras invented the bikini.... If you acutally "square" each of the sides on paper for each of the three sides of the triangle (assuming the traingle is somewhat uniform on all three sides), it will sort of look like a pair of bikini bottoms, with a woman standing on one leg! He was a weird dude, but to this day I still remember the crazy stuff he used to say...... /forums/images/graemlins/histerical.gif

Re: Quick \"Trig\" question.... [ QUOTE ] So: a^2 + b^2 = c^2 a^2 + 7^2 = 36^2 a^2 = 36^2 – 7^2 a = SQRT( 36^2 – 7^2 ) So the offset will be 36 – a 36 – 35.313 The answer is 0.687 [/ QUOTE ] /forums/images/graemlins/confused.gif /forums/images/graemlins/confused.gif /forums/images/graemlins/confused.gif

Re: Quick \"Trig\" question.... I used the same idea when I was doing my narrowing job. I wanted the rear panels to go flush with the back of the floor. Worked out well.

Re: Quick \"Trig\" question.... To big papa, and the others that don't remember trig: The a²+b²=c² thing is a triangle formula. The legs of a triangle that make a 90° angle with each other are the "a" and the "b" the connecting line is "c" the distance will equal each other as the formula above shows. There are a lot more complicated formulas that do a lot more trig tricks, but that little a²+b²=c² is the basis and most used. Russ ° = alt+0176 and ² = alt+0178 (or maybe a carrot is easier to you, just tryin to help)

Re: Quick \"Trig\" question.... LOL, yeah, I know. I just never remember the ASCII numerics and I'm too lazy to look them up. * and ^ work well enough and I don't have to try to remember the numerics... BTW, a carrot is something you eat. /forums/images/graemlins/laugh.gif I think you mean caret. /forums/images/graemlins/tongue.gif /forums/images/graemlins/rotfl.gif

Re: Quick \"Trig\" question.... Look I am eating a caret <^ /forums/images/graemlins/histerical.gif (stupid computer joke /forums/images/graemlins/doah.gif) I only remember the codes cause a lot of my classes want write ups in excel and tell us to "use the code sheet you have" stupid computer thanx to spell check I can't even write anymore. Oh back to fabwork all of you, no going off on pointless tangents. (tangent relates to the topic right now so I can write this can't I Russ??) (noted sarcasim)

Re: Quick \"Trig\" question.... /forums/images/graemlins/eek.gif Oooohhhh! That was so bad it HURT... /forums/images/graemlins/rotfl.gif

Re: Quick \"Trig\" question.... [ QUOTE ] Look I am eating a caret <^ /forums/images/graemlins/histerical.gif (stupid computer joke /forums/images/graemlins/doah.gif) I only remember the codes cause a lot of my classes want write ups in excel and tell us to "use the code sheet you have" stupid computer thanx to spell check I can't even write anymore. Oh back to fabwork all of you, no going off on pointless tangents. (tangent relates to the topic right now so I can write this can't I Russ??) (noted sarcasim) [/ QUOTE ] Introducing Matt Fennemore.....The Prince of Painful Puns!!! /forums/images/graemlins/histerical.gif