Finding the Lowest Position of a Kinematics Mechanism: A Mathematical Challenge

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The discussion revolves around solving a mathematical problem related to the kinematics of a mechanism, specifically determining the lowest position of point B1. The user has attempted various methods, including plotting points and analyzing velocities, but has not found a satisfactory mathematical solution. They initially believed that the lowest position occurs when OA aligns with DB1, but this was proven incorrect. The user seeks a more accurate mathematical approach, considering the relationship between velocities at points D and B1, and is exploring the possibility of using geometric expressions to set the derivative of the vertical position to zero. The conversation highlights the complexity of kinematic analysis and the need for a clearer mathematical framework to solve the problem.
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Hello community. I have a task from my college and no matter how hard i tried i failed to find a solution for it. The task itself is resolvable (by drawing) but i can't figure the mathematical way to solve it. And it bothers me so bad.

Homework Statement


So i have this kinematics mechanism. At the OA lever i have a known constant velocity, let's say - ω with direction counter clockwise around point O. I have given these dimensions: OA, AB, AD, DB1 and h aswell. My task is to determine the lowest position of element/point B1. Of course i plotted enough points and determined that (see the picture)... but there should be mathematical way to determine this. And i can't find it.

Initialy i thought... the lowest position of B1 is when OA is alongside to DB1. But i was wrong. Then i thought that if the velocity that occurs at point D is perpendicular to DB1 then the velocity at point B1 will be equal to 0. But this solution does not match with the position from the picture either.
c61b0913e8ff7f0c54fdd1f710bc176a_610x0.jpg
 
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Is point D a pinned joint?
 
No, its a mobile joint. My bad, i fixed it:
8cdee899b4e775f1755113f269d2da1a_610x0.jpg
 
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is length B1-D less than h?
 
.
 
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No. Here are all dimensions and values i have:

OA = 0.45m
AB = 1.1m
BD = 1.6m
B1D=0.65m
h=0.5m
ωoa = 4 rad/s (counter clockwise)

I already plotted full cycle with 6 positions and by the lowest position of B1 there i plotted another 10-12 points near it, increasing the resolution. Thus i found (roughly) the lowest position of B1 shown on the picture up there. But i still think there has to be another - much more accurate and mathmatical way to solve this.

I think that if B1 is in the lowest position its velocity should be equal to 0. Therefore at point D velocity should be perpendicular to DB1:
ωoa =Voa . OA We have ωoa and OA so Voa is known. If we separate vector Voa by sinα.Voa and cosα.Voa we will have at point B - ωbd=(sinα.Voa)/AB and the point D this velocity will be vector Vda=ωbd.BD=(sinα.Voa).(BD/AB) plus the other vector cosα.Voa. But when i calculate and plot them it doesn't seems to be perpendicular at all. And i can't figure it out why is that.
f20adbd01f3bb89d589b2961bc8cf21c_610x0.jpg
 
I think your approach looks valid. However, might a simpler be simply geometrical? Can you express y in terms of θ? Then set dy/dθ=0.
 
I didn't understand you. What are y and θ?
 
y is the vertical position of point B1 and theta is the angle OA makes with the x-axis.
 

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