Sawtooth diagram for automotive use

Dear readers,

I want to make a sawtooth diagram from an Audi RS6.

It has to look like this: https://www.google.com/search?q=sawtooth+diagram+car&rlz=1C1GCEA_enNL813NL813&sxsrf=ALeKk02PfuJadFLj3STeE6kN1-3CDSnyBw:1621332810777&source=lnms&tbm=isch&sa=X&ved=2ahUKEwiXsf6x_9LwAhVmAmMBHeZQBwUQ_AUoAXoECAEQAw&biw=1920&bih=937#imgrc=ujQRhOOBBstP-M

I have no idea where to start, I only have read something about the command sawtooth.

Can someone help me? It is very much appreciated!

Answer 1

[Scott MacKenzie]

Here's some code to create a sawtooth pattern:

samples = 100; % per cycle
cycles = 5; 
peak = 15;

y = linspace(peak,0,samples);
y = repmat(y,1,cycles);
plot(y);

[Helmar Feberwee]

I now have this, but it isn't right yet

v_max = 326.8548/3.6; %maximum vehicle speed [m/s]
r_dyn = 0.363; %dynamic rolling radius [m]
i_final = 3.204; %gear ratio final [-]
n_Pmax = 6000/60; %engine speed at maximum power [rps]

i1 = 4.714;
i2 = 3.143;
i3 = 2.106;
i4 = 1.667;
i5 = 1.285;
i6 = 1;
i7 = 0.839;
i8 = 0.667;

speed = 0:1:v_max;
n_motor_i = 0:1:v_max;

for i = 1:v_max
    n_motor_i(i) = speed(i)/r_dyn*i1*i_final;
    if n_motor_i(i)>n_Pmax
        n_motor_i(i) = speed(i)/r_dyn*i2*i_final;
        if n_motor_i(i)>n_Pmax
            n_motor_i(i) = speed(i)/r_dyn*i3*i_final;
            if n_motor_i(i)>n_Pmax
                n_motor_i(i) = speed(i)/r_dyn*i4*i_final;
                if n_motor_i(i)>n_Pmax
                    n_motor_i(i) = speed(i)/r_dyn*i5*i_final;
                    if n_motor_i(i)>n_Pmax
                        n_motor_i(i) = speed(i)/r_dyn*i6*i_final;
                        if n_motor_i(i)>n_Pmax
                            n_motor_i(i) = speed(i)/r_dyn*i7*i_final;
                            if n_motor_i(i)>n_Pmax
                                n_motor_i(i) = speed(i)/r_dyn*i8*i_final;
                            end
                        end
                    end
                end
            end
        end
    end
end
figure(1)
plot(speed,n_motor_i)

[Scott MacKenzie]

This if-tree is a bit dizzying. Plus, it can't be right since all the if-conditions are identical and occur in each iteration of the for-loop: n_motor_i(i)>n_Pmax. You need to re-think the logic. Good luck.

[Helmar Feberwee]

Hello Scott,

I now came to this script:

close all, clear all, clc

v_max = 384.159/3.6; %maximum vehicle speed (calculated) [m/s]
r_dyn = 0.363; %dynamic rolling radius [m]
i_final = 3.204; %gear ratio final [-]
n_Pmax = 6000; %engine speed at maximum power [rpm]
w_Pmax = 2*pi*6000/60; %engine speed at maximum power [rad/s]

i1 = 4.714;
i2 = 3.143;
i3 = 2.106;
i4 = 1.667;
i5 = 1.285;
i6 = 1;
i7 = 0.839;
i8 = 0.667;

speed = 0:1:v_max;
w_motor_i = 0:1:v_max;

for g = 1:v_max
    w_motor_i(g) = speed(g)/r_dyn*i1*i_final;
    if w_motor_i(g) > w_Pmax
        w_motor_i(g) = speed(g)/r_dyn*i2*i_final;
        if w_motor_i(g) > w_Pmax
            w_motor_i(g) = speed(g)/r_dyn*i3*i_final;
            if w_motor_i(g) > w_Pmax
                w_motor_i(g) = speed(g)/r_dyn*i4*i_final;
                if w_motor_i(g) > w_Pmax
                    w_motor_i(g) = speed(g)/r_dyn*i5*i_final;
                    if w_motor_i(g) > w_Pmax
                        w_motor_i(g) = speed(g)/r_dyn*i6*i_final;
                        if w_motor_i(g) > w_Pmax
                            w_motor_i(g) = speed(g)/r_dyn*i7*i_final;
                            if w_motor_i(g) > w_Pmax
                                w_motor_i(g) = speed(g)/r_dyn*i8*i_final;
                            end
                        end
                    end
                end
            end
        end
    end
end

figure(1)
plot(speed,w_motor_i)
title('Sawtooth diagram')
yline(2*pi*6000/60,'k','Pmax');
xlabel('Vehicle speed [m/s]')
ylabel('Engine speed [rad/s]')
xlim([0, v8])

This also looks like how I drew it, except for that this graph doesn't reach the maximum engines speed power line.

And by the way, thank you for your help and responses!