I have a question on vector mathematics. I have tried to find the answer to my question, but there are so many formulae and so many examples I've read through that don't apply to my question. It's lazy, I know, but I suspect if I ask here I'll get an answer quicker than spending the rest of the afternoon reading through a lot of examples that don't apply. Thanks in advance for any help.

So, the problem. There are two points (P1 and P2) in 3D space with known co-ordinates. A line with length x runs along the vector between points P1 and P2. Calculate the 3D co-ordinates of the end of the line with length x (the end of the bold line in the attached image).

Quote from speedway :

P1=(x1,y1,z1)
P2=(x2,y2,z2)

[(P2-p1)/|p1p2|]*x is what your looking for

Thanks Can you tell me the name of that equation if you know it, please?

Quote from speedway :

edited m first post

i dont think theres a name for this equation, that was basically just some logic thinking, and i cant even promise its right

Ok, thanks. I'll have a little play about with it and let you know if it's working for me.

edit: Ok, I see your edit about adding P1.

Quote from speedway :

edited m first post

i dont think theres a name for this equation, that was basically just some logic thinking, and i cant even promise its right

EDIT: i was wrong, [(P2-p1)/|p1p2|]*x is the relativ vektor when you start at P1, you therefore have to add P1 to [(P2-p1)/|p1p2|]*x

Nope, I'm having problems. I don't know whether it's my maths failing or there's a problem with the equation above.

Just to clarify the equation, to calculate the x co-ordinate of the point at the end of the line of length x is:

((x2 - x1) / (Square root ((x1*x1) + (x2*x2)))) * x

I'm making some progress (thanks for putting up with my idiocy), but there's something that doesn't seem quite right just now. I'm going to get a few hours sleep, I might do better with fresh eyes.

Bit of a bump. I was really tired and hungover when I created this thread. I was actually trying to calculate the wrong thing (so the equation given might have been perfect but it appeared to give me the wrong results). After thinking about it some more I've now clarified what I'm really trying to calculate.

A point P has known co-ordinates and a known direction vector. Perpendicular from the direction of point P there are two points Px and Py. These points have unknown co-ordinates (that's what the aim is to calculate). The distances from point P to Px and from P to Py are known. How do you calculate the co-ordinates of points Px and Py?

So the thick black line is your known direction vector? Is this still in 3D space or 2D?

Quote from Bob Smith :

So the thick black line is your known direction vector?

Yep.

Quote from Bob Smith :

Is this still in 3D space or 2D?

Still in 3D.

If I understand correctly:

If points Px and Py are perpendicular to the line intersecting P and you only know the distance from P to Px and Py, then none of your dimensions are constant. The possible locations for the unknown points exist on spherical paths Px and Py around point P. Without more information, the fact that Px and Py are perpendicular to the line P is meaningless.

If you were to tell me that points Px and Py were perpendicular to line P AND point P, then one dimension would be fixed and the points would exist in circular paths around point P.

I've just looked at the data for the Z component of the direction vector and they're all 0. I think that means the data can be considered to be in 2 dimensions rather than 3 (since the Z co-ordinate of points Px and Py will both be equal to the Z co-ordinate of point P)?

if its 2d
http://en.wikipedia.org/wiki/Rotation_matrix#Dimension_two
ill leave the rest to you everything you need to figure it out is already in here

That just means the vector runs perpendicular to the Z axis, similar to how y is always 0 in the line y=0. So no, I don't think that's an accurate assumption. It might tidy up the math a bit but that's all.

If we're now talking 2D all you need to do is take your known vector equation transform it through 90 degrees and then you'll have the vector equation for the line between the points P1 and P2. From there given your known co-ordinates at P and the distances X and Y you should be able to calculate the co-ords for P1 and P2.

I think

Found this, might be of some use:

http://www.ia.hiof.no/~borres/cgraph/math/twod/p-twod.html

edit:

And this :

http://www.euclideanspace.com/ ... metry/transform/index.htm

Hello there!, Mr. Blas is in need of your help and he didn't want to open a new thread, his question is...

Lets think that mister Blas is going to apply a force on the crank but he wants to know how many kilograms of pressure he's going to apply with his arm. Can you tell him how strong is his arm?

Note: He is a normal healthy person.

---------------------------------------------------------------
In a more serious sentence, I need this for a school work, I just need to know how many kilograms I can apply on a crank... I need to calculate a torque.

I did an experiment last night, I put my weighing device parallel to a wall and then I pushed onto it, the device said "8Kg", can I use this value?

Thanks!

I can't answer you this because you are Mexican and I'm, drunk.

Quote from Blas89 :

In a more serious sentence, I need this for a school work, I just need to know how many kilograms I can apply on a crank... I need to calculate a torque.

I did an experiment last night, I put my weighing device parallel to a wall and then I pushed onto it, the device said "8Kg", can I use this value?
Thanks!

Can u draw a better picture? Cause this is completley unclear.. I guess the maximum force you could exert with your hand would be either to push down on something with ur complete body weight or pull down it the same way..

Yeah, that's like 140kg.

I don't think i can draw something better... It's pretty simple, you're there in front of a crank and you want to move it by pushing it with your arm... Watch the new draw

Oh, ok. In that case, how much do you weigh and what are your shoes made of (and what's the floor ur standing on made of)? And how high is this crank from the ground ur standing on.

80kg and... plastic... concrete... and 1metters up the ground.

Isn't that too much info? haha, I think it would be unaccurate to calculate the force with all of those values because my mechanical device is more likely to be tested with a different altitude and maybe by a different person but if I can get an aproximate value that'll be okay!, thanks mister!

Well height actually doesn't matter

If you have 80 kg (800 N) and your shoes are rubber bottoms standing on concrete (friction coefficient 0.8) then the maximum horizontal force you can produce is what friction force between your shoes and concrete can support - and that is 0.8*800 = 640 N (or around 64 kg, but force should never be expressed in kg, and since we don't know the surface area onto which ur pressing there's no need to involve pressure).

Now if you want to know the torque you are producing on that wheel/crank, just multiply 640 N with the radius of the wheel in meters. If the wheel is 0.3 m in radius that gives you 192 Nm of torque.

Again, this is only for the case when your crank handle is in the upmost position and you are pushing it ONLY with horizontal force, as the crank rotates and let's say comes to a 7 o'clock position you horizontal force would be very low and the vertical vector component would be the only force acting - then your max force would be your weight and the max torque would be 240 Nm. Basically over the 360° degrees of crank rotation your max torque would be a sinus function.

Oh nice nice! =D, thanks to both of you, my english was the suck explaining the idea xD but thanks for trying

Actually, I will use the Kilograms and convert them to Lb, my teacher wants the torque units like that.. Lb-In

This is the drawing of your case "simplified". 800 N is your weight, F(AH) is the horizontal reaction in point A, F(BV) is the vertical reaction in point B (your shoes), F(FR) is the friction force which will be F(FR)=F(BV)*miu where miu is 0.8 (coeffiecient of friction between your shoes and concrete).

Oh thanks... I took a class about this things but I kinda fail, I would've worked it out if I opened my notebook but it would take me longer than it took you.

Thanks again

This is the mechanism scipy