A usage coverage-based approach for assessing product family design

Abstract

Computation techniques have provided designers with deeper understanding of the market niches that were neglected before. Usage contextual information has been studied in marketing research since the last century; however, little research in design engineering focuses on it. Therefore, in this paper, we analyzed the relations between usage context information and the design of products. A usage coverage model is established to integrate users and their expected usage scenarios into product family assessment. We map the user’s individual capacity together with a given product into the usage context space. The overlapping between the required usage and feasible usage can be measured. Based on this mechanism, several usage coverage indices are proposed to assess the compliance of a given product family to the expected set of usage scenarios to be covered. The original method is demonstrated on a scale-based product family of jigsaws in a redesign context. Constraint programming technique is applied to solve the physics-based causal loops that determine usage performances in a set-based design approach. Designers can rely on the results to eliminate redundant units in the family or modify the configuration of each product. The contribution of the paper is to provide an inter-disciplinary point of view to assessing the composition and configuration of a product family design.

Appendix

1.1 Jigsaw parameterization

See Fig. 11.

Fig. 11

Jigsaw parameterization: dimensions, forces, torques, and speeds

1.2 Intermediate variables for the jigsaw modeling

 

Intermediate variables (forces, speed)
ρ w (kg/m3)	Wood density
μ (no unit)	Friction factor between steel and wood
F t (N)	Translation force
F t max (N)	Maximal translation force that can be delivered by the given user defined by C s vector
F p (N)	Pressure force
F p max (N)	Maximal pressure force that can be delivered by the given user defined by C s vector
M w (N m)	Wrist torque
M w max (N m)	Maximal wrist torque that can be delivered by the given user defined by C s vector
F r (N)	Reaction force between slider and wood
F f (N)	Friction force between slider and wood
F a (N)	Advance force
$F_{{{\text{a}}_{i} }}$ (N)	Elementary advance force on tooth i
F c (N)	Cut force
$F_{{{\text{c}}_{i} }}$ (N)	Elementary cut force on tooth i
H d (m)	Height of scobs
N m	Mean number of teeth cutting wood at any moment
S c (m/s)	Mean cutting speed
S c(t) (m/s)	Instantaneous cutting speed
S a(t) (m/s)	Instantaneous advance speed

1.3 Other relations for the jigsaw physics

1.3.1 Geometrical relations

In the upper position of the blade, the upper tooth is always above the slider and is expressed by:

$$O_{\text{t}} > 0 $$

(16)

The lower tooth in the upper position is below the lower wood stick plane (teeth are cutting all along the thickness at any moment of the cutting period):

$$ t \in \left[ {\frac{1 + 2n}{2f},\frac{n + 1}{f}} \right],\quad n \in N \Rightarrow L_{\rm t} - O_{\text{t}} \ge T_{\text{c}} $$

(17)

The useful length of the blade L _t, is proportional to step s and the number of teeth n − 1:

$$ L_{\rm t} = (n - 1) \cdot s $$

(18)

1.3.2 Static relations

All the forces are non-negative (by construction on Fig. 11) and the slider always touches the wood surface. Then:

$$ F_{\text{r}} \ge 0,\quad F_{\text{a}} \ge 0,\quad F_{\text{f}} \ge 0 $$

(19)

The forces F _t and F _p delivered by the user onto the jigsaw tool can be modeled as intervals of possible values, which are bounded by the maximum allowable translation force F _t max and pressure force F _p max a user can afford:

$$ F_{\text{t}} \in [0,F_{{{\text{t\,max}}}} ],\quad F_{\text{p}} \in [0,F_{{{\text{p\,max}}}} ] $$

(20)

As previously seen in Table 1, these maximal allowable bounds are dependent on the user gender and the user skill.

The horizontal force equilibrium (advance acceleration is neglected) is expressed by:

$$ F_{\text{a}} = F_{\text{t}} - F_{\text{f}} $$

(21)

In the following, we assume that F _t may be considered as a constant entry force, imposed by the user.

The friction force between the jigsaw slider and the wood is expressed by:

$$ F_{\text{f}} = F_{\text{r}} \cdot \mu $$

(22)

The vertical force equilibrium is expressed by:

$$ F_{\text{p}} + m g + F_{\text{c}} = F_{\text{r}} $$

(24)

The momentum equilibrium at the wrist position provides the expression of the wrist torque:

$$ M_{\text{w}} = F_{\text{a}} \left( {H_{\text{w}} + \frac{{T_{\text{c}} }}{2}} \right) + F_{\text{c}} L_{\text{w}} + F_{\text{f}} H_{\text{w}} - (d + L_{\text{w}} )F_{\text{r}} $$

(25)

In the same manner, the torque in the user wrist M _w is bounded by a maximal allowable torque the user may support expressed in Table 1.

$$ M_{\text{w}} \le M_{{{\text{w\,max}}}} $$

(26)

The position of the reaction force of the slider must stay inside the slider surface area so as to avoid the jigsaw tool from flipping:

$$ d \in [O_{\text{s}} - L_{\text{s}} ,O_{\text{s}} ] $$

(27)

1.3.3 Cutting technological relations

The minimal teeth number for a successful cut is 3:

$$ T_{\text{c}} > 3s $$

(28)

The mean number of teeth cutting wood at any moment is expressed by:

$$ N_{\text{m}} = \frac{{T_{\text{c}} }}{s} $$

(29)

The elementary advance force $F_{{\text{a}}_{i}}$ on a tooth i is provided by:

$$ F_{{{\text{a}}_{i} }} = \frac{{F_{\text{a}} }}{{N_{\text{m}} }} $$

(30)

This equation means that $F_{{\text{a}}_{i}}$ is constant. During the cutting phase, when the blade is ascending, a cutting relation exists between the elementary advance force F _a(t) on a tooth i and the scob’s height H _d(t). As $F_{{\text{a}}_{i}}$ is constant, H _d is constant when the blade ascends and the relation is:

$$ t \in \left[ {\frac{1 + 2n}{2f},\frac{n + 1}{f}} \right],\quad n \in N \Rightarrow F_{{{\text{a}}_{i} }} = W_{\text{t}} \rho_{\text{w}} (H_{\text{d}} A_{2} + A_{2} ){\text{e}}^{{A_{1} (35 - \alpha )}} $$

(31)

This equation was found based on experimental measures of H _d for different types of wood and teeth (i.e., different values of W _t, ρ _W, H _d, α). The three A _i coefficients have been found experimentally:

$$ A_{1} = 0.01,\quad A_{2} = 272{,}000,\quad A_{3} = 0 $$

(32)

Finally, we obtain from (31) and (32) the following relation between H _d and $F_{{\text{a}}_{i}}$:

$$ H_{\text{d}} = \frac{{F_{{{\text{a}}_{i} }} }}{{272{,}000W_{\text{t}} \rho_{\text{w}} {\text{e}}^{0.01(35 - \alpha )} }} $$

(33)

As well, a constraint should be:

$$ H_{\text{d}} \le s $$

(34)

During the descending phase, the scob’s height is zero:

$$ t \in \left[ {\frac{n}{f},\frac{1 + 2n}{2f}} \right],\quad n \in N \Rightarrow H_{\text{d}} (t) = 0 $$

(35)

The elementary cutting force on a tooth during the cutting—ascending—phase is:

$$ F_{{{\text{c}}_{i} }} = W_{\text{t}} \rho_{\text{w}} (60{,}000 H_{\text{d}} + 20){\text{e}}^{0.01(35 - \alpha )} $$

(36)

This equation has also been found experimentally. In the same manner, the cutting force is not dependent on time.

During the descending phase, the cutting force is primarily due to the friction between the teeth and wood:

$$ F_{{{\text{c}}_{i} }} = - F_{{{\text{a}}_{i} }} \cdot \mu $$

(37)

The resulting cut force is given by:

$$ F_{\text{c}} = F_{{{\text{c}}_{i} }} \cdot N_{m} $$

(38)

1.3.4 Kinematic relations

The mean cutting speed for a forth and back of the blade (i.e., A) is:

$$ S_{\text{c}} = 2Af $$

(39)

The instantaneous cutting speed has a sinusoidal form:

$$ S_{\text{c}} (t) = 2Af\sin (2\pi ft) $$

(40)

It can be inferred that the maximal cutting speed is:

$$ S_{{{\text{c\,max}}}} = 2\pi Af $$

(41)

The instantaneous advance speed during the descending phase is null:

$$ t \in \left[ {\frac{n}{f},\frac{1 + 2n}{2f}} \right],\quad n \in N \Rightarrow S_{\text{a}} (t) = 0 $$

(42)

The instantaneous advance speed during the cutting—descending—phase is (see similar triangles on Fig. 11):

$$ t \in \left[ {\frac{1 + 2n}{2f},\frac{n + 1}{f}} \right],\quad n \in N \Rightarrow S_{\text{a}} (t) = - {\frac{H_{d} \, S_{c} \left ( t \right )}{s}} $$

(43)

The mean advance speed during one time period is:

During

$$ t \in \left[ {\frac{n}{f},\frac{1 + 2n}{2f}} \right], $$

(44)

we get

$$ S_{\text{a}} = \frac{{2H_{\text{d}} fA}}{s} $$

1.3.5 Power relations

During the cutting phase, the power provided by the engine and the hand must be enough to cut the wood, then:

$$ P_{\text{m}} + (F_{\text{t}} - F_{\text{f}} - F_{\text{a}} ) \cdot S_{\text{a}} (t) \ge F_{\text{c}} (t)S_{\text{c}} (t) $$

(45)

Given that F _t − F _f − F _a = 0 (advance acceleration neglected), then we obtain:

$$ P_{\text{m}} \ge F_{\text{c}} (t)S_{\text{c}} (t) = F_{\text{c}} S_{\text{c}} (t) $$

Consequently,

$$ P_{\text{m}} \ge F_{\text{c}} { \max }(S_{\text{c}} (t)) $$

leading to:

$$ P_{\text{m}} \ge F_{\text{c}} \cdot 2\pi Af $$

(46)

Fig. 12

Causal relations between the variables of the usage coverage modeling issue of a jigsaw