chord length

[Kenny]

Hello,

I am a first timer here, please be kind.

How do I figure out the area of a semi-circle were the hieght of the chord length is continuly changing?

Regards,

Ken

 

[tkhunny]

We're always kind. Why would we be anything else?

You're going to have to find a better translation. That one makes very little sense.

 

[Kenny]

We're always kind. Why would we be anything else? Thank you.

If you had a semi-circular trough and the water level was constantly changing. You know the depth from the bottom of the trough after the change. How would calculate the chord length in order to find the area? With this I can figure out the volume.

Does that make any sense?

Ken

 

[pappus]

We're always kind. Why would we be anything else? Thank you.

If you had a semi-circular trough and the water level was constantly changing. You know the depth from the bottom of the trough after the change. How would calculate the chord length in order to find the area? With this I can figure out the volume.

Does that make any sense?

Ken

1. Draw a sketch (see attachment)

2. Use the right triangle with the hypotenuse r, the leg (r - d) and the leg \(\displaystyle \tfrac12 c\) to determine the length of c.

 

[pka]

1. Draw a sketch (see attachment)

2. Use the right triangle with the hypotenuse r, the leg (r - d) and the leg \(\displaystyle \tfrac12 c\) to determine the length of c.

Look at fromula #8 on this page.

 

[Kenny]

Thank you Pappus and pka.
Ken

 

[Kenny]

If the radius = 2’, isn’t the radius and center angle the same in a 180 drgee semicircle?
Ref: http://mathworld.wolfram.com/CircularSegment.html

 

[pka]

If the radius = 2’, isn’t the radius and center angle the same in a 180 drgee semicircle?
Ref: http://mathworld.wolfram.com/CircularSegment.html

You do not need to know any angles.
From formula #9, the cord length is \(\displaystyle 2\sqrt{4d-d^2}\).
At that website \(\displaystyle R=2~\&~h=d\) in your diagram,

 

[Kenny]

You do not need to know any angles.
From formula #9, the cord length is \(\displaystyle 2\sqrt{4d-d^2}\).
At that website \(\displaystyle R=2~\&~h=d\) in your diagram,

Once again thank you. I am not a young man and that old saying "If you don't use it you lose it" is so true. I tried to search for formula #9 just to see what else may be there, the search was in vain. I am a hobbyist and want to incorporate this formula into some code on an MCU, it' a type of microprocessor. If I follow you all I need is the formula from #9 and this is all I need to calculate the area of liquid?

 

[pka]

Once again thank you. I am not a young man and that old saying "If you don't use it you lose it" is so true. I tried to search for formula #9 just to see what else may be there, the search was in vain. I am a hobbyist and want to incorporate this formula into some code on an MCU, it' a type of microprocessor. If I follow you all I need is the formula from #9 and this is all I need to calculate the area of liquid?

I have no idea what you mean by "tried to search for formula #9"

On this page there is a subsection entitled "cord length".
There are four formulae for cord length (6),(7),(8),&(9).
There is formula #9.

 

[Kenny]

Now I understand. I feel like an idiot.

 

[Kenny]

Could someone please double check me on this ?

Could any one double check me on this ?

Using pappus drawing: d=12, r=24
Area of a circle-circle =



= 192pi-144sqrt(3) = an area of 353.7704731993219

REF: http://mathworld.wolfram.com/Quarter-TankProblem.html