View Full Version : trig question - help me find X

nakman

03-19-2010, 10:11 PM

I am trying to draw out the bolt pattern of a 6 on 5.5 wheel spacer.. and my limited CAD ability has revealed a very rusty trig ability. When I was 16 I could have done this on a napkin in about 2 minutes... but tonight I find myself scratching my head.. :o

Can someone solve this? 30-60-90 triangle, the long leg that's not the hypotenuse is 2.25, aka the radius of the 5.5" circle. I need to know how far "up" or down, or sideways to draw, so that I can draw in the hypotenuse then create a lug hole where that intersects the circle. Here's a chance to prove how smart you is.. what's X? :confused:

wesintl

03-19-2010, 10:20 PM

1 * √2.25 ? I need to find the trig and geometry symbols but i think you will know what i'm talking about

AxleIke

03-19-2010, 10:21 PM

Sure thing:

Tan 30 = x/2.250

.5773 = x/ 2.25

x= 1.299"

fubuki

03-19-2010, 10:23 PM

http://www.thealarmclock.com/euro/images/findX.gif

AxleIke

03-19-2010, 10:26 PM

An easy way to remember these trig functions is the following pnuemonic:

Chief SOH CAH TOA (pronounced Cheif soak-a toe-a)

The SOH CAH TOA part is as follows

SINE = OPPOSITE over HYPOTENUSE

COSINE = ADJACENT over HYPOTENUSE

TANGENT = OPPOSITE over ADJACENT

Where the hypotenuse is obvious, and the adjacent side is not the hypotenuse, but the other connecting side to the angle you have, and the opposite is the one not touching the angle, if that makes any sense.

nakman

03-19-2010, 10:32 PM

Sure thing:

Tan 30 = x/2.250

.5773 = x/ 2.25

x= 1.299"

Dr. Axle, thanks man! :bowdown:

edit: yeah I remember the SOH CAH TOA from high school, just couldn't seem to get it, and using the computer calculator probably didn't help me either. I remembered the sin of a 30 degree angle was .5, yet for some reason my answer wasn't double or half of my known value. I can't wait until Gavin learns this stuff, so I can learn it all over again. :)

wesintl

03-19-2010, 10:37 PM

Doh.. I think i was in the ball park.. lol

can the airplane take off on that triangle Tim?

nakman

03-19-2010, 10:40 PM

can the airplane take off on that triangle Tim?

Only if the hypotenuse is a conveyor belt. :D

edit: and good one Fubuki, was hoping someone would post that one. :)

rover67

03-19-2010, 10:40 PM

Isaac is right but I think you are looking for a different number... since the lugs on that pattern are 60* apart it's Tan (60) = x / 2.25

so:

x = 3.897

Edit: Nevermind, you two confused me. Don't look at these numbers. :D

nakman

03-19-2010, 10:51 PM

Isaac is right but I think you are looking for a different number... since the lugs on that pattern are 60* apart it's Tan (60) = x / 2.25

so:

x = 3.897

Edit: Nevermind, you two confused me. Don't look at these numbers. :D

dude you're cracking me up..

but hey i drew a wheel spacer!

nakman

03-19-2010, 10:57 PM

Actually Marco you had it right, just had the equation flipped. I got it now..

tan60 = 2.25/x

1.732 = 2.25/x

1.732x = 2.25

x = 2.25/1.732

x = 1.299

Man it's all coming back now, I'm never going to fall asleep tonight.. how'd that quadratic formula go again? :bolt:

rover67

03-19-2010, 11:02 PM

Actually Marco you had it right, just had the equation flipped. I got it now..

tan60 = 2.25/x

1.732 = 2.25/x

1.732x = 2.25

x = 2.25/1.732

x = 1.299

Man it's all coming back now, I'm never going to fall asleep tonight.. how'd that quadratic formula go again? :bolt:

I thought you were drawing it a little different and I am too slow to realize they are both the same anyways...

Then after I posted I saw how you had it sketched.... I thought you were starting with a horizontal line...

too many :beer:s

60wag

03-20-2010, 07:26 AM

You aren't making plastic wheel spacers, are you?

nakman

03-20-2010, 08:17 AM

You aren't making plastic wheel spacers, are you?

as a means to an end, yes. as the final product, probably not.

I am annoyed with wheel shaking on these Chinese spacers I've got. The holes are too big, and there are slots that make the thing not balanced. It's tough to get it in just the right sweet spot, I had them installed well but when I took them off to put the tires on the 80 all of that is shot. I'm hopeful a more balanced wheel spacer will make this better.

But I thought a lot about what kind of compression nylon 12 will see when a lug is torqued to 85 pounds.. once it's smashed in there how much further could it go when a wheel sees side load?

ScaldedDog

03-20-2010, 08:47 AM

The easiest PC mnemonics to remember these are:

Slice Of Ham (s=o/h)

Top Of Apple (t=o/a)

Coat And Hat (c=a/h)

The one you will *never* forget, that my 8th grade algebra teacher taught me, involves what Sally can tell about Oscar, but folks would have a stroke if I posted it here. :eek:

Mark

60wag

03-20-2010, 08:54 AM

I'd be surprized if a slightly off center aluminum wheel spacer would create enough of an imbalance that it'd cause vibration. I think it's more likely that the wheel spacer is allowing the wheel to mount up off center, or out of square. Is the wheel located by the studs, or by the hub in the center? I wouldn't run a plastic experiment at high speed as I bet the lug nuts won't stay tight.

How about adding some rings (sleeves) to the existing spacers to tighten up the bolt holes?

nakman

03-20-2010, 09:08 AM

The wheel is centered by the studs, by the shanks on the lug nuts, and they're not long enough to penetrate the spacers so I think they're "falling down" behind the wheel. Yeah some little sleeves is another approach, that doesn't resolve the weight imbalance issue but that is likely negligible.

I ran each spacer across some emery paper to remove any high spots.. don't know if that did anything. They seem flat, but I haven't measured.

FJBRADY

03-20-2010, 09:24 AM

http://www.thealarmclock.com/euro/images/findX.gif

:D

Just so all of you know Trig is the reason I am a sales guy......I am very happy there are about 100 engineer's in the group to help Nakman out. :drumsticks:

subzali

03-21-2010, 08:47 AM

(I'm going way back here)

Quadratic function: f(x)=ax^2+bx+c. To find the roots, do the following:

(-b +/- sqroot(b^2-4ac))/2a (?)

BTW, the plane takes off. I proved it with physics.

http://forum.ih8mud.com/chit-chat-section/69513-physics-question-riddle-good-one-read-5.html

TIMZTOY

03-21-2010, 08:14 PM

:rolleyes:

AxleIke

03-21-2010, 10:59 PM

(I'm going way back here)

Quadratic function: f(x)=ax^2+bx+c. To find the roots, do the following:

(-b +/- sqroot(b^2-4ac))/2a (?)

BTW, the plane takes off. I proved it with physics.

http://forum.ih8mud.com/chit-chat-section/69513-physics-question-riddle-good-one-read-5.html

Nice Matt! FWIW, I never understood the issue with the plane taking off. Anyone with common sense can figure that out.

Good work on the explanation though! Just completely confused on why anyone would find that hard to understand, and why it lasted for that many pages? :confused:

Also, as a side note, you did remember correctly both a quadratic function, and the quadratic formula (going off of the fact that there were question marks in parenthesis)

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