# How to calculate spray cone angle?

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RPR le 15 Fév 2024
Modifié(e) : Matt J le 15 Fév 2024
Hi everyone,
I am doing experiments and capturing the spray pattern using high speed camera. I need to measure spray angle, diameter distribution and velocity of the spray. I have attached one of the images from 1000 images of spray pattern. For calculating spray angle, I have used the matlab code mentioned in this link https://in.mathworks.com/matlabcentral/answers/2003472-how-to-calculate-spray-cone-angle . But I could not get the correct spray angle. Could anyone provide the code to measure spray angle. I have attached the results which I have obtained.
And also, with the images is it possible to obtain diameter distribution of the particles and the velocity of the sprayed particles. If it is possible, please provide code for the same.
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Matt J le 15 Fév 2024
Modifié(e) : Matt J le 15 Fév 2024
Please give additional examples of spray images where the code fails, so we can get a sense of the general task and its difficulties.
RPR le 15 Fév 2024
thank you for your answer. I have attached 10 more spray images for your reference.

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### Réponses (2)

Image Analyst le 15 Fév 2024
First of all, before doing any image processing, you need to get better pictures. There is not enough contrast between your spray and your background. I suggest trying using a black velvet backdrop for your background. Then use focused/colimated lighting from both sides to illuminate the spray, or use non-colimated light but have baffles in place so as to not illuminate the background. Once you have good images, it should be fairly straightforward.
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Matt J le 15 Fév 2024
Modifié(e) : Matt J le 15 Fév 2024
BW = imfill( bwareafilt(imclose(entropyfilt(im2gray(Image))>4.5,strel('disk',5)),1),'holes');
BW=bwareafilt(~imclose(~BW,ones(1,35)),1);
imshow(labeloverlay(Image,BW,'Colormap','hot','Transp',0.7));
BW=BW(1:floor(height(BW)/2),:);
BW=BW(any(BW,2),:); N=height(BW);
[~,J1]=max(BW,[],2);
[~,J2]=min(cummax(BW,2,'reverse'),[],2);
p1=polyfit(1:N,J1,1);
p2=polyfit(1:N,J2,1);
angle=abs(atand(p1(1))) + abs(atand(p2(1)))
angle = 74.8004
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RPR le 15 Fév 2024
Déplacé(e) : Matt J le 15 Fév 2024
Sir, I am getting same answer (74.8004) for spray angle for all operating conditions. could you please help me.
Matt J le 15 Fév 2024
Modifié(e) : Matt J le 15 Fév 2024
All your images look the same to me, so it comes as no surprise that the angles are all the same.

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