Help understanding the role of ϕ in equations for C_limb, COL_walk

Hello,

Thank you in advance for your help. I’m hoping to get some help understanding the role of ϕ in the equation below. Essentially, does having a larger ϕ make COL_walk larger or smaller?

. . . . .\(\displaystyle \mbox{C}_{\mbox{limb}}\, =\, 2kBfgM_L D\phi \, \big|1\, -\, T^2\, T_0^{-2}\big|\)

. . . . .\(\displaystyle \mbox{COL}_{\mbox{walk}}\, =\, 8kfU^2L^{-1}\, \big[1\, +\, \cos\left(\frac{\phi}{2}\right)\big]^{-1}\, \big[1\, +\, \tan\left(\frac{\phi}{2}\right)\big]\, +\, \mbox{C}_{\mbox{limb}}\)

Thank you so much,

Michael

mgranatosky said:

....does having a larger ϕ make COL_walk larger or smaller?

. . . . .\(\displaystyle \mbox{C}_{\mbox{limb}}\, =\, 2kBfgM_L D\phi \, \big|1\, -\, T^2\, T_0^{-2}\big|\)

. . . . .\(\displaystyle \mbox{COL}_{\mbox{walk}}\, =\, 8kfU^2L^{-1}\, \big[1\, +\, \cos\left(\frac{\phi}{2}\right)\big]^{-1}\, \big[1\, +\, \tan\left(\frac{\phi}{2}\right)\big]\, +\, \mbox{C}_{\mbox{limb}}\)

Click to expand...

What did you find, when you plugged in larger values?

Thank you for your response. Unfortunately, my math skills are not where I would like them to be. Would you mind working out an example assuming that all other variables are held constant?

mgranatosky said:

Thank you for your response. Unfortunately, my math skills are not where I would like them to be. Would you mind working out an example assuming that all other variables are held constant?

Click to expand...

It's just plugging a number in, and simplifying.

Copy the "C_limb" expression to the end of the COL_walk expression. Then pick a value for phi. I'm guessing that the values for phi are supposed to be angle measures, probably in radians (since it would make sense to multiply b, f, D, etc, by "degrees", which aren't actually numbers). But if this is related to walking or climbing, then there may be restrictions on the values which can be input. For instance, people don't usually walk slopes of, say, 120 degrees (so they'd have to be hanging upside-down like a gecko).

Please reply with the other necessary information (context, instructions, etc), and show what you have tried so far. When you plugged things into your calculator, what did you get? And so forth, so we can try to figure out what the difficulty is. Thank you!

Okay, let me see if my math works out. If you assume that all the other variables are 1, then Climb is negligible because will [FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]T[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]T[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Main]∣[/FONT]result in zero.

So, you're limited to the bottom equation with Climb.

If you assume that one angle is 25, then you get:

(8*1**1*1² *1 ) * (1 + cos (25/2))^-1 * (1 + tan (25/2))

8 * (1 + 0.97629601) ^-1 * (1+ 0.22169466)

8 * (0.50599) * (1.22167) = 4.945

If you assume a second angle is 50, then you get:

(8*1**1*1² *1 ) * (1 + cos (50/2))^-1 * (1 + tan (50/2))

8 * (1 + 0.90630779) ^-1 * (1+ 0.46630766)

8 * (0.5245742609) * (1.46630766) = 6.1534

So, larger angles would result in higher COLwalk?

Let me know if/where I messed up.

All the best,

Michael




[FONT=MathJax_Main][/FONT]

Just out of curiosity, do you have the ability to graph this function out and see how it varies with phi?

It seems like it's just of the form

\(\displaystyle A\left[1 + \cos\left(\frac{\phi}{2}\right)\right]^{-1}\left[1 + \tan\left(\frac{\phi}{2}\right)\right] + B\phi\)

Where I've condensed all the multiplicative factors that aren't phi into single constants

I attempted to work this out in excel. Is this correct?

mgranatosky said:

I attempted to work this out in excel. Is this correct?

Click to expand...

As far as I can tell, what you have is not correct. There are a number of issues

1) Rather than picking A = 10 and B = 10 specifically, if you're going to work it out numerically using Excel anyway, you could actually just plug in all the values for the constants that make up A and B in the original problem. However, since you're just looking to figure out the shape of the function for now, to see how it varies with phi, picking arbitrary values like A = 10 and B = 10 is ok as a start.

2) The shape of your function looks wrong. It's monotonic (which means steadily increasing or decreasing). That is very suspicious, since both cosine and tangent are periodic functions, so I would expect your answer to have some sort of cyclic variation (or at least some sort of variation up and down). I suspect the problem is one of units. Phi seems like it is supposed to physically represent some sort of angle. And, based on the fact that you are plugging in values for phi like 90 or 100, I'm guessing you are choosing phi to be in degrees. But you have to remember that an angle in degrees is never what goes into a trigonometric function. Plugging in an angle in degrees into cosine or tangent is nonsense. Those functions take dimensionless numbers, which means that the angles you plug into them have to be in radians. I doubt very much that you intended to plug in 50 radians or 100 radians, since radian values are redundant outside the range of 0 to 2*pi. Convert your angles to radians first.

3) There is also the B*phi term, which is not a trig function. What units of phi is it expecting to be plugged into it? Well, it's unclear. It depends on the units of all the constants that go into it. If all those constants have units of C_limb, then phi needs to be a dimensionless number, and therefore should be in radians. This seems like the most likely case to me. However, if all those constants multiply together to produce something that has units of C_limb per degree, then the expression is obviously expecting phi in degrees. The point is, you have to figure that out based on your knowledge of the physical constants.

4) Even if I plug in phi in degrees into the expression (which I want to emphasize, is the wrong thing to do), I don't get the same numerical values as you, which leads me to believe you might have an error in your Excel formulas. I get these numbers here:

10 [ 81.45528447]
15 [ 177.5205456]
20 [ 302.42816687]
25 [ 254.67280289]
30 [ 305.99248256]
35 [ 321.74127835]
40 [ 422.98985998]
45 [ 572.96044153]
50 [ 504.35150848]
55 [ 550.76767057]
60 [ 553.17024555]
65 [ 669.68816588]
70 [ 853.03171688]
75 [ 754.03089215]
80 [ 796.4806867]
85 [ 755.63966519]
90 [ 917.17522733]
95 [ 1149.5590293]
100 [ 1003.70540446]
105 [ 1042.74292698]

Note, also, that this wrong function is not monotonic either. it has lots of negative spikes. it just happens to always be positive for these input values.