Stuck solving a complex equation for T

In summary, the equation for predicting an athlete's finishing time is F(T)=-1.04*w*C*D/T+k(D/T)^3+2(1+w/W)*(D/T)^2*1/T, where g(T)=-0.0023. Bisection is used to find the value of T in the middle where F(T)=-1.04 and g(T)=-0.0023. This value is then used to predict the finishing time.
  • #1
CallmeSam
4
0
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
 
Last edited:
Mathematics news on Phys.org
  • #2
CallmeSam said:
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.
 
  • #3
I like Serena said:
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
 
Last edited:
  • #4
CallmeSam said:
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.
 
  • #5
I like Serena said:
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.
 
  • #6
CallmeSam said:
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 Tlow = 0.
And we need an upper boundary Thigh such that g(Thigh) has an opposite sign with respect to g(Tlow), so that we can be sure that the solution will be in between.
From there we can apply bisection.
 
  • #7
CallmeSam said:
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.
 
  • #8
Theia said:
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.
 

FAQ: Stuck solving a complex equation for T

1. What is a complex equation?

A complex equation is a mathematical expression that contains variables, constants, and mathematical operations such as addition, subtraction, multiplication, and division. It often involves multiple steps and requires a systematic approach to solve.

2. Why is it difficult to solve a complex equation for T?

Complex equations can be difficult to solve for T because they often involve multiple variables and require a series of steps to find the solution. Additionally, some equations may have more than one possible solution or may require the use of advanced techniques such as calculus.

3. What is the best approach for solving a complex equation for T?

The best approach for solving a complex equation for T is to break it down into smaller, manageable steps. Start by identifying the variables and constants in the equation and then use algebraic techniques to isolate T on one side of the equation. If necessary, you can also use graphing or numerical methods to find the solution.

4. Can technology be used to solve complex equations for T?

Yes, technology such as calculators and computer software can be very helpful in solving complex equations for T. These tools can perform calculations quickly and accurately, and some even have built-in functions specifically designed for solving equations.

5. How can I check if my solution for T is correct?

You can check your solution for T by plugging it back into the original equation and evaluating both sides. If they are equal, then your solution is correct. Additionally, you can use technology or ask a colleague to verify your solution.

Similar threads

Replies
2
Views
1K
Replies
11
Views
2K
Replies
1
Views
849
Replies
1
Views
1K
Replies
3
Views
7K
Back
Top