In the Introduction section of Ian Millingtons book "Game Physics Engine Development", he assumes that you have a certain level of mathematical knowledge. An example of what you are expected to understand is that x = 4 / t * sin(theta)^2 is the same as t = 4 / x * sin(theta)^2. I have tried to work this out but I cannot seem to resolve it at the last step. Here are my workings:

• x = 4 / t * sin(theta)^2
• t * x = 4 / t * t * sin(theta)^2 - (Multiply both sides by t to bring t over to the LHS)
• t * x = 4 * sin(theta)^2 - (The ts on the RHS cancel out)
• t * x / x = 4 * sin(theta)^2 / x - (Now divide both sides by x to remove the x from the LHS)
• t = 4 * sin(theta)^2 / x - (This is what I have ended up with)

What I have ended up with is the whole of the right hand side of the equation over x. Ian Millington ends with only the 4 being divided by x. What am I misunderstanding here? My current understanding is that what is done to one side of the equation must be done to the other in order for it to remain true. Therefore, if all of the left hand side is divided by x then all of the right hand side must be divided by x, not just 4. If this is true, then does that not mean that sin(theta)^2 must also be divided by x? If so, how can this be resolved? Thanks.

Apologies for the formatting, I cannot figure out how to use the formatter.

Ah, sorry you're right that's not very clear, it should be:

x = (4 / t) * sin(theta)^2 is the same as t = (4 / x) * sin(theta)^2

• x = (4 / t) * sin(theta)^2
• t * x = (4 / t) * t * sin(theta)^2 - (Multiply both sides by t to bring t over to the LHS)
• t * x = 4 * sin(theta)^2 - (The ts on the RHS cancel out)
• (t * x) / x = (4 * sin(theta)^2) / x - (Now divide both sides by x to remove the x from the LHS)
• t = (4 * sin(theta)^2) / x - (This is what I have ended up with)

18 hours ago, swiftcoder said:

t = (4*sin(theta)^2) / x is equivalent to  t = (4/x)*sin(theta)^2

Would you mind explaining how? Obviously I'm wrong but to my mind t = (4 * sin(theta)^2) / x is the same as t = (4 / x) * (sin(theta)^2 / x), since both 4 and sin(theta)^2 are divided by the x.

18 hours ago, swiftcoder said:

Keep in mind that 4/x == 4 * (1/x)

This makes sense to me, but I'm not making the logical connection between this and the equation.

18 hours ago, swiftcoder said:

multiplication is associative, so you can reorder multiplied terms at will

I don't understand. Would you mind explaining further?

Maths is not my strong point. It was just recently that I learned how to solve for variables in an equation and to re-arrange equations. It's incredible really, I feel like I have unlocked so many doors with this knowledge. No longer is it a mystery to me how expressions like "y = sin(theta) * radius" are arrived at. I had always understood through SOHCAHTOA that sin(theta) = y / radius, but I never understood how it was that people could re-arrange that to the equation above. It really is a great feeling when you make logical connections in your brain and finally understand what you did not before. Hopefully you can help me to make the connections to demystify my mystified brain on this issue.

Okay, I understand reordering terms when multiplying, I just didn't know what you meant by associative. However, I don't understand how you got from:

t = (4*sin(theta)^2) / x 

to

t = 4 * sin(theta)^2 * (1 / x)

Let's just take the right hand side. I understand that 4 / x is equal to 4 * (1 / x) because:

4 * (1 / x) = (4 / 1) * (1 / x) = 4 * 1 / 1 * x => 4 / x

However, the right hand side of the equation isn't 4 / x, if it was it would already be solved. sin(theta)^2 is part of the numerator. This must be where my knowledge fails, because as I understand it everything in the numerator must be divided by the denominator. This would mean that both 4 and sin(theta)^2 must each be divided by x. Am I incorrect in thinking that? Can you pick and choose which term in a set of terms in the numerator to divide by the denominator? 

On 3/6/2018 at 11:54 PM, Seer said:

x = 4 / t * sin(theta)^2  ==  t = 4 / x * sin(theta)^2.

Substation is a easy way that anyone can use to solve this problem.

a = b/c * D  ->  c = b/a * D Substitute every thing for a variable.

a = 1/2 * 3  Substitute all variables for simple numbers except the a.

1/2 * 3 = 1.5 Solve a. 

1.5 = 1/2 *3 -> 2 = 1/1.5 * 3

Now lets try your way.

1.) 1.5 * 2  =  (1/2*2) * 3  "(Multiply both sides by t to bring t over to the LHS)"

2.) 1.5 * 2 = 1*3 " (The ts on the RHS cancel out)"

3.) (1.5 * 2) / 1.5 = (1*3)/ 1.5   "(Now divide both sides by x to remove the x from the LHS)"

4.) 2 = (1*3)/ 1.5  "(This is what I have ended up with)"

Correct! You where never wrong you just stopped too early.

5.) 2 = (1/1.5) * 3 See Solved. or  t= (4/x) * sin(theta)^2

On 3/6/2018 at 11:54 PM, Seer said:

What I have ended up with is the whole of the right hand side of the equation over x. My current understanding is that what is done to one side of the equation 

On 3/7/2018 at 12:38 AM, Seer said:

t * x = (4 / t) * t * sin(theta)^2 - (Multiply both sides by t to bring t over to the LHS)

The same thing happens in your first step.

(x) *t = ((4/t)* sin(theta)^2) *t - (Multiply both sides by t to bring t over to the LHS)

multiplication and division , BODMAS