Is there a way to solve the transcendental equation O=R{1-Cos(28.65S/R)} for R?

  • Thread starter Thread starter damskippy
  • Start date Start date
  • Tags Tags
    Formula Stuck
damskippy
Messages
2
Reaction score
0
Hey all,
I need some help rearranging for R.

O=R{1-Cos(28.65S/R)}

For those curious, its a Horizontal Sightline Offset formula used in geometric road design.

thanks
 
Mathematics news on Phys.org
Is that the letter O or zero to the left of the equals?

If zero then can't you just divide by R and solve what's left?
 
If it's not zero then that's the first time I've ever seen a variable being represented by the letter O. It would also mean that you can't explicitly solve for R.
 
Apologies for the confusion, it's an O for Offset. This should make it a little clearer, (HSO being Horizontal Sightline Offset).

HSO=R{1-Cos(28.65S/R)}

But what your saying, you can't solve the above equation for R. Might explain why I couldn't, apart from my poor maths. Thanks for your help.
 
damskippy said:
But what your saying, you can't solve the above equation for R. Might explain why I couldn't, apart from my poor maths. Thanks for your help.

Because we can't describe the solution to the above equation in terms of any finite number of algebraic operations (which include +,-,*,/,powers,roots).

For example, even though it's well known that the solution to ex=2 is x=ln(2), this is a transcendental. We can't actually find the value of ln(2) exactly (althought it's irrational anyway) in terms of a finite number of algebraic operations. This function ln(x) was created and given a name because it's so commonly used.

The taylor series (which is an expansion of a transcendetal function with an infinite amount of algebraic operations to describe it) for the exponential is

e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+...

So I believe if a computer were to approximate the value of e3 for example, it would truncate the taylor series for ex and then plug in x=3 and solve that.

Similarly, your equation is transcendental, but the difference here is that there isn't any commonly known function to describe the solution. Surely someone could've named the solution whatever they liked, but it would still need to be solved numerically like the ln(x) function.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top