anyone know the correct measurements fot dex? how do you measuere out 80 grams?
anyone know the correct measurements fot dex? how do you measuere out 80 grams?
doesnt it come with a scooper?
im not sure but i think 1 tblsp=5g
i take my protein scooper and do double the amount of dextrose than protein... easy enough for me. i put 2 scoop protein, 4 scoop dextrose. (this is using 100% whey which is 23g for 29g protein so it measures out easily)
xfade7
my dex is 4 grams sugar per 1.25 teaspoon and i am assuming that we are talking 80 grams sugar when we talk 2:1 ratio
yeah thats a good call, thanks alot, i appreciate the help
1/2 cup is 75 grams according to swolecat i believe. What i did was i took 1.25 tsp per 4 grams and multiplied it by 20 to get 80 grams. thats 25 teaspoons total. divided approximately by 3 tbls per teaspoon gives you 8 tablespoons. I dumped 8 tablespoons into a 1/2 measuring cup and its about dead on. you cannot go by doubling the scooper of your protein for sure. I have 1 100% whey protein that is a 20 grams a scoop and is 1/2 the size of my other protein which 25 grams a scoop. One is more dense of a powder than the other.
Originally Posted by xfade7
i think u are grossly miscalculating how much dextrose you are adding to your shake. 2 scoops of optimum whey = 46g protein and 2 scoops of dextrose USING the optimum whey scooper = about 75-80g or 1/2 cup.
so since you are adding 4 scoops you are intaking 160g of carbs.. bad idea.
Originally Posted by HiFi
My Optimum Nutrition whey package says each scoop is 28g and contains 23g of protein, therefore, you could do 2 scoops whey(46g pro), and 3 scoops of dex(84g carb) to roughly get the 2:1 ratio.
to be sure just look at how much sugar is in each serving. then go by that
my dextrose from supp direct is 75g for 1/3 cup
Wrong, dextrose is a lot more dense than protein powder. I just weighed 1 level optimum nutrition scoop of dex and it comes out to 48.7g. So, at 3 scoops you're getting nearly twice what you thought.Originally Posted by HammerCurler
There are currently 1 users browsing this thread. (0 members and 1 guests)