Open longzw001116 opened 1 year ago
In solver.py line 75-78:
image_width = params['f'] * np.tan(20.0 * np.pi / 180.0) * np.sqrt(2)
image_width = image_width * params['magnification'] / params['sensor_pixel']
params['image_width'] = np.int(2*dim * np.ceil(image_width / (2*dim) ))
I think i know why np.sqrt(2). The FOV=40° maybe refer to diagonal FOV.
In solver.py line 75-78:
image_width = params['f'] * np.tan(20.0 * np.pi / 180.0) * np.sqrt(2) image_width = image_width * params['magnification'] / params['sensor_pixel'] params['image_width'] = np.int(2*dim * np.ceil(image_width / (2*dim) ))
- why do we times the np.sqrt(2)? In my opinion, it should be 2.
- why params['magnification'] = 8.1 instead of 10 indicated in Supplementary Figure 7?
Since thess is no description about the best loss weighting coefficients {λ1, λperc, λgrad} used in training, I wonder whether it is the value set in the ”run_train.sh“ that λ1=1.0, λperc=0.1 and λgrad=0.0.
Hi, thanks for your interest in our work and apologies for the late reply. I recommend playing around with the loss weighting as I did not find any particular setting to be a clear "best". Nevertheless, I recommend putting most of the weight into λ1 and putting lesser weight into the other coefficients.
I think i know why np.sqrt(2). The FOV=40° maybe refer to diagonal FOV.
In solver.py line 75-78:
image_width = params['f'] * np.tan(20.0 * np.pi / 180.0) * np.sqrt(2) image_width = image_width * params['magnification'] / params['sensor_pixel'] params['image_width'] = np.int(2*dim * np.ceil(image_width / (2*dim) ))
- why do we times the np.sqrt(2)? In my opinion, it should be 2.
- why params['magnification'] = 8.1 instead of 10 indicated in Supplementary Figure 7?
Since thess is no description about the best loss weighting coefficients {λ1, λperc, λgrad} used in training, I wonder whether it is the value set in the ”run_train.sh“ that λ1=1.0, λperc=0.1 and λgrad=0.0.