I'm a total rookie at calculus and this holds me back from getting to the bottom of many problems like this :( that's why I probably misunderstood your solution.
But to me it looks like this: α is an unknown constant -- constant yes, but yet unknown, undetermined; as such the recipe misses one...
Thank you very much for your time. Sorry for the late reply. Reading your post and "constant of integration" on Wikipedia raised some intriguing facts:
- A can be decomposed into terms of v and t making the formula simpler.
- v(t) can't be really determined as α isn't determined in the first...
Thank you very much. It looks like what I'm asking for. Surprisingly as it may seem -- all the calculus I know is raw formulae for derivatives like (3x4)' = 12x3, sin(x)' = cos(x) etc. So this drastically outweighs my mathematical knowledge.
If I'm not asking too much, could you please, write an...
Thanks for replying.
I need to find both At (acceleration) and vt which, problematically, are interdependent: A depends on v as per given formula (At = vt2 * 0.04), v depends on A by Newton's laws. We ignore gravity. I don't know why you brought force into discussion.
I tried something on the...
Hello.
I've seen on Wikipedia (http://en.wikipedia.org/wiki/Airsoft_pellets#pellet_ballistics) a formula to find the aerodynamic drag (velocity dependent negative acceleration) for a spherical body. Knowing body's mass, diameter and air density I simplified it to the form:
A = vt2 * 0.04
where A...
At a second thought, the 2nd part isn't such a big issue. I could simply estimate a rough linear-logarithmic function; the user won't notice too much. The precision isn't crucial.
LE:
I know I sound fawning but I totally respect hard work and hard mind. I realize that I'm 2566 posts away from...
Thank you so much. You already did very much, Please, don't mind the starting point. And with the second part, don't waste too much of your time. I feel in debt already
LE:
Oops! too late.
Thank you very much. Man that was fast. What if I change start and end points?
I can't relate them to the function...
Now that I read your reply I realize that the 2nd part is... a separate function by itself.
f(D) = n
OFFTOPIC:
I saw your sig in another thread and couldn't ask there. So I do...
My mathematically profane answer: The points of the second line are bigger :smile:
If you say for any point of short(?) line there exist correspond a point on the long(?) line, it means that the points of the latter are the projections of the shorter line's points. He he.
Hello.
Long time, no see. From all the forums I'm a member of, I feel this is the most strange to me as my math is very basic. So, here is what I'd be thankful to be helped with:
I'm making a "mechanic" that I'm going to use in game and software developing for tons of times if I'll pull it...
Thanks a lot! After asiduous searching on the web, I realized that the soft I use to make the game has arcsin already implemented. Doh!
If you sometime feel like doing simple games, go try Game Maker; it's friendly. :)
almost there
I found the formula for the value:
R=sqrt(V1²+V2²+2*V1*V2*cos(α))
As for the direction, I'm stuck at:
sin(β)=V2*sin(α)/R
I need β; how can I pull out β in this equation?
The very formula of parallelogram vectors merge?
Ok, forget anything else!
Anywayz, I have 2 vectors; I know the directions and the values for both; what's the formula for finding the resulting merged vector direction and value? (the vectors aren't perpendicular)
I need an equation...
Thank you; it's hard to apply this in what I'm making, though; I basically want to know the value of a 2nd vector (v2) please take a look at my scheme, cause I got the feeling v1 not equals v2 (v2<v1).
So, how can I determine how much of the tank force is applied to the truck?
If v2 is a...
I tried at "Intro physics" and was warn that my post might be deleted if not stuck to a certain template. So, here I am.
I'm an outsider in physics, yet I'm trying to make a 2d game with collisions.
I know that I can "split" a vector into two and combine two into one by this rule; I need the...