Dual Firing the Torpedo: Ideas

 Dual Firing the torpedo

	Connect to one axle, out of phase by 90 degrees
	Potentially change this linkage to a CAM (would require spring to actively push wedge…)

Initial Torpedo Spring Calculations

Feb 23 2023

Worked backwards from a desired distance making the following assumptions:

• We want the torpedo to travel 0.5m or about 20” 0.2m or about 8”

• Torpedo mass of 10g

• Compression of the spring by 1 cm or 0.01m

• Damping/drag coefficient of 0.1

• Sources indicate a streamline shape has a drag coefficient of 0.05; to compensate for it being in water, having fins and other factors I doubled the number

Calculations redone in next table

Worked backwards from the speed of the torpedo after launch to find desired speed

Wrote out ODE of the spring in launching action

Did not solve on my calculator as the notes suggest—was giving a domain error. Instead, I solved the system on MATLAB to find the minimum spring constant with the desired velocity.

% Lindsay Wright % RoboSubSpring.m % State Space for Spring and Water Damping % Last Modified: 2/23/23 %Some assumptions: a drag coefficient of 0.01 Ns/m, pushback of 1 cm at the %start t=0.001:0.001:0.1; kPlot=zeros(1,length(0.005:0.005:0.5)); vPlot=zeros(1,length(0.005:0.005:0.5)); for i=1:length(t)     kPlot(i)=((pi/2/t(i))^2+25)/100;     vPlot(i)=(pi/2)*0.01/t(i)*exp(-5*t(i)); end figure plot(t,kPlot) xlabel('launch time (s)') ylabel('Spring Constant (N/m)') figure plot(t,vPlot) xlabel('launch time (s)') ylabel('Velocity (m/s)') table=zeros(3,length(t)); table(1,:)=t; table(2,:)=kPlot; table(3,:)=vPlot; disp(table)

Conclusion: Need a minimum spring constant of 2750 N/m That is a powerful spring!

Changing the distance requirement to 0.2m (about 8”) makes the needed initial velocity 2 m/s This yields a spring constant of ~500 N/m

Changing the distance requirement to 0.3m (about 1’) makes the needed initial velocity 3 m/s This yields a spring constant of ~1000 N/m or ~560 lbs/in

close all; clear; clc;

% Lindsay Wright

% RoboSubSpring.m

% State Space for Spring and Water Damping

% Last Modified: 2/25/23

%Some assumptions: a drag coefficient of 0.01 Ns/m, pushback of 1 cm at the

%start

t=0.001:0.001:0.1;

kPlot=zeros(1,length(0.005:0.005:0.5));

vPlot=zeros(1,length(0.005:0.005:0.5));

for i=1:length(t)

   kPlot(i)=((pi/2/t(i))^2+25)/100;

   vPlot(i)=(pi/2)*0.03/t(i)*exp(-5*t(i)); %changed 0.01 kg to 0.03 kg

end

desired_travel = 5; % in meters

diff_val = abs(desired_travel*ones(1,length(t))-vPlot);

index = find(min(diff_val) == diff_val);

%index = find(vPlot == );

figure

hold on

plot(t,kPlot)

plot(t(index),kPlot(index),'LineStyle', 'none','Marker','o','MarkerSize',5,'MarkerFaceColor','red','MarkerEdgeColor','none')

hold off

xlabel('launch time (s)')

ylabel('Spring Constant (N/m)')

figure

hold on

plot(t,vPlot)

plot(t(index),vPlot(index),'LineStyle', 'none','Marker','o','MarkerSize',5,'MarkerFaceColor','red','MarkerEdgeColor','none')

hold off

xlabel('launch time (s)')

ylabel('Velocity (m/s)')

table=zeros(3,length(t));

table(1,:)=t;

table(2,:)=kPlot;

table(3,:)=vPlot;

%disp(table)

fprintf('need: %07.4f N/m or %08.4f lb/in to travel: %07.4f meters\n',kPlot(index),kPlot(index)*8.8507457676,vPlot(index))

20230301

Calculations redone to account for differences in mass. Also realized I made an error in calculating the roots of the characteristic equation, yielding a smaller spring constant than expected.

% Lindsay Wright % RoboSubSpring.m % State Space for Spring and Water Damping % Last Modified: 3/1/23 %Some assumptions: a drag coefficient of 0.01 Ns/m, pushback of 1 cm at the %start t=0.001:0.001:0.1; kPlot=zeros(1,length(0.005:0.005:0.5)); vPlot=zeros(1,length(0.005:0.005:0.5)); for i=1:length(t)   kPlot(i)=((pi/2/t(i))^2+25/9)*3/100;   vPlot(i)=(pi/2)*0.01/t(i); %changed 0.01 kg to 0.03 kg end desired_travel = 0.3; % in meters diff_val = abs(desired_travel*ones(1,length(t))-vPlot); index = find(min(diff_val) == diff_val); %index = find(vPlot == ); figure hold on plot(t,kPlot) plot(t(index),kPlot(index),'LineStyle', 'none','Marker','o','MarkerSize',5,'MarkerFaceColor','red','MarkerEdgeColor','none') hold off xlabel('launch time (s)') ylabel('Spring Constant (N/m)') figure hold on plot(t,vPlot) plot(t(index),vPlot(index),'LineStyle', 'none','Marker','o','MarkerSize',5,'MarkerFaceColor','red','MarkerEdgeColor','none') hold off xlabel('launch time (s)') ylabel('Velocity (m/s)') table=zeros(3,length(t)); table(1,:)=t; table(2,:)=kPlot; table(3,:)=vPlot; %disp(table) fprintf('need: %07.4f N/m or %08.4f lb/in to travel: %07.4f meters\n',kPlot(index),kPlot(index)*8.8507457676,vPlot(index))

Or, ~1.89 lb/in