Methodologies for weight estimation of fixed and flapping wing micro air vehicles

Abstract

One of the important steps in the sizing process of fixed and flapping wing micro air vehicles (MAVs) is weight estimation of the electrical and structural components. In order to enhance the flight performance and endurance of MAVs, it is required to carefully estimate their weight with a minimum error. In this study, methodologies to estimate the weight of fixed and flapping wing MAVs are proposed. After dividing the total weight of the MAV into weights of structural and electrical components, these two weights are separately identified. The weight of the MAV electrical components is estimated by using engineering design techniques and the weight of the structure is identified by using statistical and computational methods. The proposed methodology for structural weight estimation is based on calculating the percentage of the used material in the construction of different parts of MAVs and then presenting the weight of each part in terms of the wing surface. The proposed computational method gives the exact estimation for the weight of each structure component, such as wing, tail, fuselage, and etc. Based on the offered method for weight estimation of MAVs, the weight estimation of a fixed wing MAV with inverse Zimmerman planform and a flapping wing MAV named “Thunder I” are experimentally shown. This developed methodology gives guidelines for weight estimation and determination of the structural weight percentages in order to design and fabricate efficient fixed and flapping wing MAVs.

Appendix

Substituting Eq. (6) into Eq. (7), one obtains:

$$t_{wing} = t_{MAC} = 0.06MAC = 0.06 \times \frac{2}{3}c_{r} \left( {\frac{{1 + \lambda + \lambda^{2} }}{1 + \lambda }} \right)$$

The parameter c _r can be written as a function of the wing surface as follows:

$$S_{Wing} = (c_{r} + c_{t} )\frac{b}{2} = c_{r} \left( {1 + \lambda } \right)\frac{{\sqrt {AR \times S_{wing} } }}{2} \to c_{r} = \frac{2}{{\left( {1 + \lambda } \right)}}\sqrt {\frac{{S_{wing} }}{AR}}$$

Then, by substituting the value of c _r , in Eq. (7), we get:

$$t_{wing} = t_{MAC} = 0.06MAC = 0.06 \times \frac{2}{3}\frac{2}{{\left( {1 + \lambda } \right)}}\sqrt {\frac{{S_{wing} }}{AR}} \left( {\frac{{1 + \lambda + \lambda^{2} }}{1 + \lambda }} \right) = 0.08\sqrt {\frac{{S_{wing} }}{AR}} \left( {\frac{{1 + \lambda + \lambda^{2} }}{{1 + 2\lambda + \lambda^{2} }}} \right)$$

If we consider a delta planform with λ > 0.375 and using Eq. (16), the tail arm can be expressed as:

$$l = c_{r} - x_{N} - 0.1\bar{c}$$

Substituting Eq. (15) into Eq. (6), and using the calculated formula for c _r , one gets:

$$l = c_{r} - x_{N} - 0.1\bar{c} = c_{r} - \frac{{c_{r} }}{4} - \frac{b}{6}\frac{1 + 2\lambda }{1 + \lambda }\tan_{{\Phi_{0.25} }} - 0.1 \times \frac{2}{3}c_{r} \left( {\frac{{1 + \lambda + \lambda^{2} }}{1 + \lambda }} \right)$$

Using the previous expression and Fig. 7, tanΦ _0.25 can be written as:

$$\tan_{{\Phi_{0.25} }} = \frac{{c_{r} - 0.25c_{r} - (c_{t} - 0.25c_{t} )}}{b/2} = \frac{{0.75(c_{r} - c_{t} )}}{b/2} = \frac{{1.5(c_{r} - c_{t} )}}{b} = \frac{{1.5c_{r} (1 - \lambda )}}{b}$$

Substituting tanΦ _0.25 in the expression of l, one obtains:

$$l = c_{r} - \frac{{c_{r} }}{4} - \frac{b}{6}\frac{1 + 2\lambda }{1 + \lambda } \times \frac{{1.5c_{r} (1 - \lambda )}}{b} - 0.1 \times \frac{2}{3}c_{r} \left( {\frac{{1 + \lambda + \lambda^{2} }}{1 + \lambda }} \right)$$

The previous expression can be simplified to be:

$$l = c_{r} \left[ {0.75 - 0.25\left( {\frac{{1 + \lambda - 2\lambda^{2} }}{1 + \lambda }} \right) - \frac{0.2}{3}\left( {\frac{{1 + \lambda + \lambda^{2} }}{1 + \lambda }} \right)} \right]$$

Substituting c _r by its expression, we have:

$$l = \frac{2}{{\left( {1 + \lambda } \right)}}\sqrt {\frac{{S_{wing} }}{AR}} \left[ {0.75 - 0.25\left( {\frac{{1 + \lambda - 2\lambda^{2} }}{1 + \lambda }} \right) - \frac{0.2}{3}\left( {\frac{{1 + \lambda + \lambda^{2} }}{1 + \lambda }} \right)} \right]$$

Simplifying the previous expression, one gets:

$$l = \sqrt {\frac{{S_{wing} }}{AR}} \frac{2}{{\left( {1 + \lambda } \right)}}\left[ {\frac{{2.25(1 + \lambda ) - 0.75(1 + \lambda - 2\lambda^{2} ) - 0.2(1 + \lambda + \lambda^{2} )}}{3(1 + \lambda )}} \right]$$

And then:

$$l = \left[ {\frac{{2.6(1 + \lambda + \lambda^{2} )}}{{3(1 + \lambda )^{2} \sqrt {AR} }}} \right]\sqrt {S_{wing} }$$

As for Eq. (71), we have:

$$W_{f} = 0.45\rho_{f} t_{f} \left( {c_{r} + 0.1\sqrt {AR \times S_{wing} } } \right)\sqrt {AR \times S_{wing} }$$

Substituting c _r by its expression, we have:

$$W_{f} = 0.45\rho_{f} t_{f} \left( {\frac{2}{{\left( {1 + \lambda } \right)}}\sqrt {\frac{{S_{wing} }}{AR}} + 0.1\sqrt {AR \times S_{wing} } } \right)\sqrt {AR \times S_{wing} }$$

Then,

$$W_{f} = 0.45\rho_{f} t_{f} \left( {\frac{{2S_{wing} }}{{\left( {\lambda + 1} \right)}} + 0.1\left( {AR \times S_{wing} } \right)} \right)$$

(71)

According to Tables 11 and 14, for leading edge spar, we have:

$$W_{{LE{\text{-}}Spars}} = \frac{\pi }{4}\rho_{Ls} D_{Ls}^{2} \sqrt {AR \times S_{wing} }$$

Substituting the corresponding values for the density and the diameter of the carbon rod and aspect ratio, one obtains:

$$W_{{LE{\text{-}}Spars}} = \frac{\pi }{4}\rho_{Ls} D_{Ls}^{2} \sqrt {AR \times S_{wing} } = \frac{\pi }{4} \times 1400 \times 0.005^{2} \times \sqrt {3.85 \times S_{wing} } \approx 0.04\sqrt {S_{wing} }$$

$$W_{{Diagonal{\text{-}}Spars}} = \frac{\pi }{2}\rho_{ds} D_{ds}^{2} \sqrt {\frac{{AR \times S_{wing} }}{4} + c_{r}^{2} } = \frac{\pi }{2} \times 1400 \times 0.0025^{2} \sqrt {\frac{{3.85 \times S_{wing} }}{4} + c_{r}^{2} }$$

$$W_{{Diagonal{\text{-}}Spars}} \approx 0.014\sqrt {0.9625S_{wing} + c_{r}^{2} }$$

$$c_{r} = 0.35 \times b = 0.35\sqrt {AR \times S_{wing} } = 0.7\sqrt {S_{wing} }$$

$$W_{{Diagonal{\text{-}}Spars}} \approx 0.014\sqrt {0.9625S_{wing} + 0.49S_{wing} } \approx 0.02\sqrt {S_{wing} }$$