MHB Stuck solving a complex equation for T

[CallmeSam]

Hi I wonder if anyone can help. I am not even sure I am on the right forum.

I cannot solve this equation for t. It is the final sequence of a number of equations in a book about modelling athletic performance using bioenergetics. I had a high school maths education 40 years ago and I’m stumped. I have tried Mathway app to solve equations but I think requires a human touch... The author says itÂ’s easy to create a predicted finishing time “just solve for T”.

P= 1/F(T)[(1+w/W)F*C*D/T+k(D/T)^3+2(1+w/W)(D/T)^2*1/T]

There is a hint it may involve “bisection” whatever this may be.

Any help appreciated.

Many Thanks

Bill

 

[I like Serena]

Hi I wonder if anyone can help. I am not even sure I am on the right forum.

I cannot solve this equation for t. It is the final sequence of a number of equations in a book about modelling athletic performance using bioenergetics. I had a high school maths education 40 years ago and I’m stumped. I have tried Mathway app to solve equations but I think requires a human touch... The author says itÂ’s easy to create a predicted finishing time “just solve for T”.

P= 1/F(T)[(1+w/W)F*C*D/T+k(D/T)^3+2(1+w/W)(D/T)^2*1/T]

There is a hint it may involve “bisection” whatever this may be.

Any help appreciated.

Many Thanks

Bill

Hi CallmeSam, welcome to MHB!

First, can you clarify what F and k are?
That is, I see F(T), which suggests that it's a function, but I also see F without parentheses, which suggests that it's a constant.
The symbol k is also followed by a parenthesis, implying that it might be a function.

 

[CallmeSam]

Hi CallmeSam, welcome to MHB!

First, can you clarify what F and k are?
That is, I see F(T), which suggests that it's a function, but I also see F without parentheses, which suggests that it's a constant.
The symbol k is also followed by a parenthesis, implying that it might be a function.

Hi thanks for replying. In the equation k is constant of 0.0023. F(T) is described as a dimensionless quantity and is a function of T (time) which can be calculated independently with a range of values from 1.4 to 1.04. F in subsequent parts of the equation is actually Fg I realize dropping the g has confused it with F(T).

Does this help?

Bill

 

[I like Serena]

Hi thanks for replying. In the equation k is constant of 0.0023. F(T) is described as a dimensionless quantity and is a function of T (time) which can be calculated independently with a range of values from 1.4 to 1.04. F in subsequent parts of the equation is actually Fg I realize dropping the g has confused it with F(T).

Does this help?

Bill

Yep.
So apparently we can only evaluate the equation for $1.04 \le T \le 1.4$.
And for some such value we hope to find that it's solution.

Let's define: g(T) = 1/F(T)[(1+w/W)F*C*D/T+k(D/T)^3+2(1+w/W)(D/T)^2*1/T] - P
Then we're looking for a value of T such that g(T) = 0.

If g(1.04) and g(1.4) are on opposite sides of 0 (if one is positive then the other is negative), we can be sure that somewhere in between it must be 0.
Assuming that's the case, then this is where bisection comes in.
We pick the value in the middle, which is (1.04 + 1.4) / 2 and check if it's positive or negative.
Consequently we can tell if the solution is in the left half, or in the right half.
Now repeat until we consider the solution to be accurate enough.

 

[CallmeSam]

Yep.
So apparently we can only evaluate the equation for $1.04 \le T \le 1.4$.
And for some such value we hope to find that it's solution.

Let's define: g(T) = 1/F(T)[(1+w/W)F*C*D/T+k(D/T)^3+2(1+w/W)(D/T)^2*1/T] - P
Then we're looking for a value of T such that g(T) = 0.

If g(1.04) and g(1.4) are on opposite sides of 0 (if one is positive then the other is negative), we can be sure that somewhere in between it must be 0.
Assuming that's the case, then this is where bisection comes in.
We pick the value in the middle, which is (1.04 + 1.4) / 2 and check if it's positive or negative.
Consequently we can tell if the solution is in the left half, or in the right half.
Now repeat until we consider the solution to be accurate enough.

I may not have been clear - the value of F(T) is can be between 1.04 and 1.4... does your solution apply to that? If so I presume I replace F(T), w, W, F, C, D, P, etc with the known values and then trial values between 1.04 and 1.4 as g(T) until we get a Value for T that appears accurate enough?

Thank you for your kind help so far... I appreciate it.

 

[I like Serena]

I may not have been clear - the value of F(T) is can be between 1.04 and 1.4... does your solution apply to that? If so I presume I replace F(T), w, W, F, C, D, P, etc with the known values and then trial values between 1.04 and 1.4 as g(T) until we get a Value for T that appears accurate enough?

Thank you for your kind help so far... I appreciate it.

Ah okay.
Yes, my solution still applies.
But it does mean that we need to start with a different lower boundary and upper boundary for T.
If T is a time then an obvious choice for the lower boundary is T_low = 0.
And we need an upper boundary T_high such that g(T_high) has an opposite sign with respect to g(T_low), so that we can be sure that the solution will be in between.
From there we can apply bisection.

 

[Theia]

I may not have been clear - the value of F(T) is can be between 1.04 and 1.4... does your solution apply to that? If so I presume I replace F(T), w, W, F, C, D, P, etc with the known values and then trial values between 1.04 and 1.4 as g(T) until we get a Value for T that appears accurate enough?

A side note: It may be quite hard work to do only by hand and simple calculator. Maybe you could create an Excel/Calc sheet or even better, a computer program to reduce your work and avoid computation errors.

 

[CallmeSam]

A side note: It may be quite hard work to do only by hand and simple calculator. Maybe you could create an Excel/Calc sheet or even better, a computer program to reduce your work and avoid computation errors.

Unfortunately being able to create even the simplest computer program is beyond me

However I will look at using Calc sheet... though it may take a while. Many Thanks.