- #1
JohnnyGui
- 796
- 51
Hello,
I did a calculation to determine whether a liquid with a fixed volume ##V##, would be spread over a larger surface area ##A## on the inside mantle of a cylinder, if the cylinder has a larger radius ##r##. So I’d like to find a relationship between the radius ##r## and the area ##A## over which the liquid is spread in the cylinder. The length of the cylinder ##L## and the volume of liequid ##V## are fixed.
If the liquid with volume ##V## is put in the cylinder with a radius ##r##, it would have the following picture when cut through;
- The area ##O## is formulated as: ##\frac{r^2}{2} \cdot (θ – sin(θ)) = O##.
- The volume ##V## of the liquid would therefore be defined as: ##\frac{r^2}{2} \cdot (θ – sin(θ)) \cdot L =V##.
- The area ##A## of the inside “mantle” of the cylinder over which the liquid is spread is defined as ##L \cdot r \cdot θ = A##.
Combining these formulas will give the relationship of V, θ, r and A in the following form:
$$\frac{2V}{r(1-\frac{sin(θ)}{θ})} = A$$
The radius ##r## can be formulated in terms of θ as well since ##\sqrt{\frac{2V}{L(θ-sin(θ))}}=r##. The end formula being:
$$\frac{2V}{\sqrt{\frac{2V}{L(θ-sin(θ)}} \cdot (1-\frac{sin(θ)}{θ})} = A$$
Plotting the relationship ##\sqrt{\frac{2V}{L(θ-sin(θ))}}=r## with θ being the variable, ##V## and ##L## being fixed values, shows a peculiar graph:
There are points in the graph of which the slope is actually zero, which means that a change in θ doesn’t give a change in radius. Looking at the graph as if the radius is the variable (which is my initial question) thus says that there are values of ##r## in which the change θ is vertical, as if it means that the radius ##r## doesn’t need to change to make θ change. I’m not sure how to interpret this.
How is this possible if the length ##L## and the volume of the liquid ##V## are fixed?
I did a calculation to determine whether a liquid with a fixed volume ##V##, would be spread over a larger surface area ##A## on the inside mantle of a cylinder, if the cylinder has a larger radius ##r##. So I’d like to find a relationship between the radius ##r## and the area ##A## over which the liquid is spread in the cylinder. The length of the cylinder ##L## and the volume of liequid ##V## are fixed.
If the liquid with volume ##V## is put in the cylinder with a radius ##r##, it would have the following picture when cut through;
- The area ##O## is formulated as: ##\frac{r^2}{2} \cdot (θ – sin(θ)) = O##.
- The volume ##V## of the liquid would therefore be defined as: ##\frac{r^2}{2} \cdot (θ – sin(θ)) \cdot L =V##.
- The area ##A## of the inside “mantle” of the cylinder over which the liquid is spread is defined as ##L \cdot r \cdot θ = A##.
Combining these formulas will give the relationship of V, θ, r and A in the following form:
$$\frac{2V}{r(1-\frac{sin(θ)}{θ})} = A$$
The radius ##r## can be formulated in terms of θ as well since ##\sqrt{\frac{2V}{L(θ-sin(θ))}}=r##. The end formula being:
$$\frac{2V}{\sqrt{\frac{2V}{L(θ-sin(θ)}} \cdot (1-\frac{sin(θ)}{θ})} = A$$
Plotting the relationship ##\sqrt{\frac{2V}{L(θ-sin(θ))}}=r## with θ being the variable, ##V## and ##L## being fixed values, shows a peculiar graph:
There are points in the graph of which the slope is actually zero, which means that a change in θ doesn’t give a change in radius. Looking at the graph as if the radius is the variable (which is my initial question) thus says that there are values of ##r## in which the change θ is vertical, as if it means that the radius ##r## doesn’t need to change to make θ change. I’m not sure how to interpret this.
How is this possible if the length ##L## and the volume of the liquid ##V## are fixed?
Last edited: