Question about muscle growth

AlexEtheridge1996
AlexEtheridge1996 Posts: 65 Member
edited November 20 in Fitness and Exercise
We all know that working out is important, and nutrition is key to optimal growth. However, can somebody help me with a question.
To grow muscle, is it more important to be in a surplus of calories on the day you workout, or the days following your workout in which you recover and grow? People will say you need a surplus all the time, but which is more optimal?
I ask this because im interested in recomp and calorie cycling, but dont see how it can be optimal for muscle growth not being in a surplus all the time (workoit and recovery days), however j can see how its useful for gaining muscle and losing fat (albeit slowly)

Replies

  • senecarr
    senecarr Posts: 5,377 Member
    I don't think there is any conclusive evidence either way on calorie cycling.
    There is a pretty good and long thread on recomping here:
    http://community.myfitnesspal.com/en/discussion/comment/33095589#Comment_33095589
    You might want to shoot this question over there.
  • SonyaCele
    SonyaCele Posts: 2,841 Member
    feed the workout and feed the recovery. surplus all the time is my opinion. But if i had a limited source of food or seriously watching calories, i'd probably fuel the recovery more and the workout less. but thats just me, everyone will have their own opinion and everyone's bodies are slightly different.
  • CSARdiver
    CSARdiver Posts: 6,252 Member
    Follow senecarr's link. USMCMP started an outstanding conversation on recomp and some really great exchanges going on there.

    While there is little to no scientific evidence supporting any sort of timing - I personally inrease protein/carbs following resistance training. It takes ~48hrs for any muscle response, which corresponds to the time it takes for your body to uptake and fully metabolize the protein and carbs.

    As for surplus your body doesn't react that quickly to intake response, so for recomp a surplus of carbs/protein is optimal.
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