MHB Find the Angle of Parable Tangent from Human Running Step

AI Thread Summary
The discussion focuses on calculating the angle of the parable tangent related to human running steps, specifically the arc traced by a foot during a step. A formula provided by a researcher indicates that the stride angle tangent is calculated as 4 times the height divided by the step length. The participants express confusion about the reasoning behind multiplying the height by 4. A mathematical derivation is presented, confirming that the angle can be expressed as the arctangent of the ratio of 4 times the height to the step length. This analysis highlights the relationship between step height and length in determining the angle of the arc in human running motion.
Barkiernan
Messages
1
Reaction score
0
I am a masters student studying motion analysis in human running.

I need to find the angle of the parable tangent derived from the theoretical arc traced by a foot during a step and the ground (see attached). The arc comprises of a persons step height and step length and I need to find the angle of the arc it creates. No other research regarding this angle has manual calculated it.

I have contacted the researcher and he gave me the following formula:

Stride (step) angle tangent = 4*height / Step length
Therefore, the Stride (step) angle = tan-1(4*height/step length)”

However we are not sure why the height is multiplied by 4 ?

View attachment 9260

Thank you
 

Attachments

  • BBCFCE55-FE06-4875-BFC7-9B0B1D8EBB01.jpeg
    BBCFCE55-FE06-4875-BFC7-9B0B1D8EBB01.jpeg
    31.4 KB · Views: 130
Mathematics news on Phys.org
Hello, and welcome to MHB! (Wave)

If we let \(\ell\) be the stride length, and \(h\) be the max height, and orient our coordinate axes such that the "toe off" is at the origin, then we have:

$$f(x)=kx(x-\ell)$$

Now, we must have:

$$f\left(\frac{\ell}{2}\right)=h$$

$$k\left(\frac{\ell}{2}\right)\left(\frac{\ell}{2}-\ell\right)=h\implies k=-\frac{4h}{\ell^2}$$

And so:

$$f(x)=-\frac{4h}{\ell^2}x(x-\ell)=-\frac{4h}{\ell^2}x^2+\frac{4h}{\ell}x$$

From this we find:

$$f'(x)=-\frac{8h}{\ell^2}x+\frac{4h}{\ell}$$

And then:

$$f'(0)=\frac{4h}{\ell}$$

Thus, we may conclude:

$$\alpha=\arctan\left(\frac{4h}{\ell}\right)$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top