The curvature of the foot can be divided into true and false. The true curvature caused by the bending of the bones. Such a problem requires consultation and treatment from podiatrist. Fortunately, this situation does not occur very often.
Usually, we are talking about about the curvature of the legs. In this case, the visibility of the curvature arises because of the unfavorable location of the soft tissue. The bones of the legs straight.
To determine whether you have curvature of the legs, you must remove your shoes and stand up straight on a flat surface. Put feet together and close your thighs tightly. Perfectly straight the feet should be in contact at three points - ankles, calves and knees. This creates four "skylight" above the knee, below the knee, above the ankle and between the foot and ankle. The true curvature of the feet is diagnosed in the absence of the two points of contact.
If there is no contact between the knee and calf, it is referred to as O-shaped curvature of the bones. If there is no contact of the ankles and calves, this defect is called X-shaped curvature.
If there is no only one point of contact in the region of the calf muscle - there is reason to speak of about the curvature of the legs.
Correct about the curvature of the legs is possible by means of special exercises. If you for some reason do not fit, then come to the aid of plastic surgery. About the curvature quite easily avoided with silicone implants or injections of your own fat.
If true the curve of the foot contour is generally fruitless. The only way to fix it is to seek the assistance of an orthopedic surgeon. In such cases, the doctor may perform the surgery for excision of the bones of the lower leg. For the bones in the desired position after the operation is applied the Ilizarov apparatus. It should be noted that this operation is a major surgery and has a lot of contraindications. Besides, it exists the risk of various complications. The rehabilitation period can be 2-3 months.
The easiest and most enjoyable way to get rid of the curvature of the legs is to pick the right clothes. For example, jeans and pants straight cut visually "straighten" any curvature. Mini skirts can be combined with tights with the large geometric pattern, or with boots-bottomtime. For too slim leg fit tights with horizontal stripes.
Advice 2: How to calculate the length of the curve
When calculating any length, remember that it is a finite quantity, that is just a number. If you mean the arc length of the curve, such problem is solved by using the definite integral (planar case) or a curvilinear integral of the first kind (the arc length). The arc AB is denoted IAV.
The first case (flat). Let IAV is set to a flat curve y = f(x). The argument of the function will change in the range from a to b and it is continuously differentiable this segment. Find the length L of the arc IAV (Fig. 1A). To solve this problem, break the reporting period into elementary segments ∆xi, i=1,2,...,n. As a result, the IAV will be divided into elementary arcs ∆Ui, plots the graph of y=f(x) at each of the elementary segments. Find the length ∆Li of an elementary arc approximately, replacing it with the corresponding chord. It is possible to increase in to replace the differentials and use the Pythagorean theorem. After the issuance of the square root of the differential dx will get the result shown in figure 1b.
The second case (arc IAV set parametrically). x=x(t), y=y(t), tє[α,β]. Function x(t) and y(t) have continuous derivatives on the interval this interval. Find their differentials. dx=f’(t)dt, dy=f’(t)dt. Substitute these differentials into the formula to calculate the length of the arc in the first case. Remove dt from the square root under the integral, put x(α)=a, x(β)=b and come to the formula for calculating arc length in this case (see Fig. 2A).
The third event. Arc IAV graph of a function given in polar coordinates ρ=ρ(φ) of the Polar angle φ during the passage of the arc changes from α to β. The function ρ(φ)) has a continuous derivative on the interval of consideration. In such a situation, the easiest way to use the data obtained in the previous step. Choose φ as a parameter and substitute in equations relating polar and Cartesian coordinates x=ρcosφ y=ρsinφ. Differentiate the formula and substitute the squares of the derivatives in the expression in Fig. 2A. After a bit of identical transformations that are based mainly on the use of trigonometric identities (cos)^2+(sinφ)^2=1 we obtain the formula for arc length in polar coordinates (see Fig.2b).
The fourth case (spatial curve, defined parametrically). x=x(t), y=y(t), z=z(t), tє[α,β]. Strictly speaking, we should apply the line integral of the first kind (the arc length). Curvilinear integrals calculated by translating them in ordinary certain. As a result, the practical answer remains the same as case two, the only difference is that under the root will appear in an extension term – the square of the derivative z’(t) (see Fig. 2C).